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This is a Statue Park puzzle, made in honor of the 50th anniversary of the Apollo 11 mission.
Rules of Statue Park:
Shade some cells of the grid to form the given set of pieces. Pieces may be rotated or reflected.
Pieces cannot be orthogonally adjacent (though they can touch at a corner).
All unshaded cells must form a single connected group.
Any cells with black circles must be shaded; any cells with white circles must be unshaded.
Finally got there.
The tiles read "For Apollo 11" :)
Step by step solution.
Two squares of the same colour indicate that we know that they are going to be shaded, and covered by the same piece. Two squares of a different colour indicate that we know that they are going to be shaded, and not covered by the same piece.
Here the pink are the same colour because if not, the piece covering the lower left dot would need to be adjacent to the other one.
The two bottom right dots cannot possibly be separated by a run of connected empty squares, therefore we must have empty squares in the middle as follows (at the only two places that can connect the left and right parts of the board).
It is now impossible to avoid a string of blue as follows:
And therefore the two dots nearby have to be of the same colour, and therefore the lower right one is of a different one. This has easy consequences as indicated (the very bottom right corner is empty otherwise two tiles of different colours would be adjacent).
Now the very top dot has to be red, otherwise red has too little room to live, and then the other two dots nearby have to be red as well otherwise they have too little room to live. There are new empty squares at the bottom, where nothing could possibly fit.
Now let us wonder where a 2x2 square can be, assuming that orange is not one of them: there has to be two on the right as indicated, and one somewhere on the left. But then the orange are going to be (F shaped, so as not to touch the magenta L underneath, and therefore) adjacent to the bottom right 2x2 square, a contradiction. [Also credits to jafe in the comments for that part]
Therefore we have:
From here, one easily checks that: if Green is the bulky 7, then there can be at most one more piece on the left side (it can be at the top left or at the bottom left depending how Bulky Green is placed), and there is not enough room on the right side to place three more pieces, so this is impossible. Therefore the bulky one is Purple (there is not enough room anywhere else). Moreover, if it was not leaning towards the centre (as below), then (similarly to the above reasoning) there would not be enough room for more than one piece more on the right, and not enough room on the left for 3 pieces, a contradiction. [I've tried to find something less ugly at this step but failed]
From here, the fourth Green square has to be at the top, otherwise there would be no room for two pieces on the left (in particular an inverted L at the bottom left would disconnect the empty squares at the top-left side). With Green in such a position, there can be no 2x2 squares on the left.
From here, we disqualify one by one the squares that would disconnect the empty zones if filled, and two L shapes remain qualified, which in turn implies that the pink is F-shaped as follows.
