This is related to the mathematical field of graph theory, and your puzzle is essentially a finite-state machine. Each platform has 8 possible states, so the entire system has $8^3=512$ states ('vertices' in graph theory). Each button maps one of those 512 states to precisely another one (an arrow, or 'directed edge' in graph theory). It's not recommended to draw such a large graph, but you can imagine that following the 'button 1' edge eight times will lead to the initial vertex/state. For button 2 and 3, it's only a cycle of length four. Also, parts of the graph will be disjoint (separate) from other parts, which means you can never reach them from the initial (solved) position, and consequently if the puzzle is in such a state, you can never solve it.
I haven't looked at your puzzle in detail, but I imagine that there are only $8 \cdot 4 \cdot 4 = 128$ states reachable from the final position (the lengths of the cycles of each button); the puzzle will probably be harder if
- you make sure every button rotates every platform
- the other buttons get a cycle length of 8 as well
- the rotations are longer (e.g. 135 degrees or even 225; longer rotations are harder to visualize for the solver)
- there's some kind of maximum to the number of times you can press each button
- you can't press the same button twice in a row
On the other hand, the difficulty level of this puzzle might be perfect for your audience. You never know until your test!
Another way to look at this puzzle, which provides an easy way to solve it, is to look at modular arithmetic, in this case modulo 8. A state of the puzzle is denoted as $(p_1, p_2, p_3)$ where each number denotes the number of 45 degree turns the platforms ($p_1$ for P1, $p_2$ for P2, $p_3$ for P3) have made from the solution. So the depicted situation would be $(0, 4, 2)$. Pushing button 1 adds $(1, 1, 1)$, button 2 adds $(0, 2, 2)$ and button 3 adds $(2, 2, 0)$. Whenever you encounter a number of 8 or higher, just subtract 8. It doesn't matter in which order you push the buttons; let's say you push button 1 $b_1$ times, button 2 $b_2$ times and button 3 $b_3$ times. That means, since we want to reach $(0, 0, 0)$:
$$0 + 1b_1 + 0b_2 + 2b_3 = 0 \pmod 8$$
$$4 + 1b_1 + 2b_2 + 2b_3 = 0 \pmod 8$$
$$2 + 1b_1 + 2b_2 + 0b_3 = 0 \pmod 8$$
or equivalently
$$b_1 + \ \ \ \ \ \ \ \ \ \ 2b_3 = 0 \pmod 8$$
$$b_1 + 2b_2 + 2b_3 = 4 \pmod 8$$
$$b_1 + 2b_2 \ \ \ \ \ \ \ \ \ \ = 6 \pmod 8$$
Subtracting the first equation from the second gives $2b_2 = 4$ so you need to push button 2 two times; substituting that in the second equation gives $b_1 + 2 \cdot 2 = 6$, so $b_1 = 2$ so button 1 two times as well. Putting that in the first equation gives $2 + 2b_3 = 0 \pmod 8$ so $b_3$ must be 3 ($2 + 2 \cdot 3 = 8 = 0 \pmod 8$); you need to push button 3 three times.
