Note: I initially shelved my write-up of this puzzle when @Reinier posted their solution when I was midway through. However, after reflection - even though the two are similar in their logic up to a point, albeit with some steps in a different order - I figure my write-up may still be useful to people interested in this puzzle as there was one step in @Reinier's that I couldn't quite follow, plus I inserted the confirmed arrow clues into the puzzle graphic as I went, which I think makes it easier to comprehend... You can be the judge of that!
To break into this puzzle, let's use the E clues as an illustration of the general method...
By looking at the E cells positioned around the edges, we can see that the E clue cannot involve a Left arrow (there is an E on the left edge), a Right arrow (E on the right edge) or a Down arrow (E on the bottom edge). We are left to conclude that the E must therefore be the Up arrow clue. Inserting these into the E positions immediately allows us to blank out adjacent Left, Right and Down sides.
Of course, we can also blank out all spaces that contain a clue. And we can repeat our edge-based logical deductions to blank out all immediately adjacent spaces above an A, B or C (on top edge), below a C or I (on bottom edge), to the left of an H (on left edge) or right of an F or G (on right edge).
And now we can see that additionally:
- H cannot have a Down arrow (there are no shaded squares below the H in Column 3);
- F cannot have a Left arrow (no shaded squares left of the F in Column 3);
- G cannot have an Up arrow (no shaded squares above the G in R3C13).
No pentomino can fit in the enclosed 2x2 block in the bottom left corner, so blank this out. Looking at the C in the bottom row, this now has no shaded squares to its left, so C cannot have a Left arrow. In the previous step we have already ruled out C having Up and Down arrows too, so C must be the Right arrow clue.
Next, note that we've already ascertained that H cannot have a Left or a Down arrow. Now that the Up and Right clues have been identified (E and C, respectively), H must therefore be the the Up-and-Right clue. Some additional shading and blanking out follows.
This in turn fixes F as our Up-and-Down arrow clue.
Consider next the top-left corner of the grid.
If R2C3 is left blank, then a second P pentomino will form in the top left-hand corner area (since there must be a shaded cell above the clue in R3C1). To avoid this illegal situation, R2C3 must be shaded (and thus R4C3 also). Furthermore, at least one of R1C3 and R2C4 must be shaded, again to avoid a P pentomino forming.
This means that the A clue must contain a Left and/or a Down arrow (an additional Right arrow is still a possibility too). With this in mind, we know that one of R10C1, R11C2 and R12C2 must be shaded. Importantly, whichever of these is shaded, we would definitely end up needing to shade R11C2 for continuity. This means that the A clue definitely includes a Down arrow and we can shade R2C4 for definite.
This has a further knock-on effect in the top-left corner - since at least one of R1C1 and R2C1 must be shaded (due to the Up clue immediately below) and connected to the shaded cells in R2C3-4, it is impossible for R1C5 to be shaded - the A clue is therefore either Down or Left-and-Down (it cannot include a Right arrow - some blankings out follow, including the now isolated A1C7).
In fact, R2C5 must also be blanked out too, since if the A clue is Left-and-Down, shading this space would result in two separate pentominoes touching.
Note also at this point that it is impossible for R12C3 to be shaded - we cannot use a pentomino to satisfy the Right-Up clue in R12C1 whilst also ensuring that R12C3 and R13C4 are part of the same pentomino. There are a couple of knock-on shadings and blankings out for continuity.
Because we have now shaded R11C3, there must be at least one shaded square in R15C4-6, to satisfy the Right clue in R15C3. This means that R14C5 must be shaded as a result, which means that the B clue must include a Left arrow. This has a big knock-on effect at the top of the grid, where the B clue in row 1 has a long line of blanks to its left - equivalent blanks must now also be entered on its other available flanks.
Note that at this stage we can narrow B down to one of the Left clue, the Left-and-Down clue, or the Left-and-Right clue - it cannot be Left-Right-and-Down, or two separate pentominoes will touch illegally at the bottom of the grid.
These additional blankings-out mean we now know R3C9-11 at a minimum must all be shaded to satisfy the Up clue in row 4. This means that the D clue must include a Down arrow. We can apply this information to the D clue in row 14 and then note that D must also include a Right arrow and D cannot include a Left arrow, since R14C9 must be blank to avoid separate pentominoes touching. This means D must be the Right-and-Down clue (we know it cannot include an Up arrow due to the blank in R1C10).
We can also resolve the Up clue in the bottom row now, since we know R14C8 must be shaded (so that the shaded square upwards is closer than the one to its right).
Consider now the Up-and-Down clue in column 14. Can we shade the spaces directly above and below it? If we do then the Z pentomino is immediately placed, the B clue in row 9 then becomes a Left-and-Right arrow and the G clue in row 3 is forced to be a Left arrow. Knock-on deductions for the G in R13C13 and the B in row 14 then create two W pentominoes at the bottom - a contradiction!
Instead, this means the Up-and-Down arrow in column 14 can only be satisfied by the already-shaded square in R5C14 and its counterpart in the bottom row, which immediately places the U pentomino in that corner and forces the pentomino to the left of I to be the L shape. This then means that I is the Left-and-Right arrow clue and G is the Down arrow clue.
We can now place the Z pentomino on the right, and also conclude that the A clue must be Down-and-Left, as this is its only option remaining. In turn, the B clue can now only be the Left arrow clue and we have identified all of the clues!
This immediately allows us to resolve many more squares:
Now focus on the unresolved Up-Right clue in column 1. Because there is a shaded square two spaces below it, this must be resolved by shading its two immediately adjacent squares. This instantly gives us the W pentomino.
Next note that R5C4 cannot be shaded, or a second F pentomino will be formed. This forces the placement of the N pentomino, which in turn forces the top left pentomino to be the Y.
The T, X and V pentominoes are now all forced.
And finally, we note that only the I pentomino remains to be placed, and there is only one way to do that while satisfying the Right arrow clue in the bottom row. All other spaces are blank and the puzzle is complete at last!