If $\mathcal{P}(A)=\mathcal{P}(B)$, then $A=B$? Prove, disprove, or give a counterexample:
  
  If $\mathcal{P}(A)=\mathcal{P}(B)$, then $A=B$.


Assume $\mathcal{P}(A)=\mathcal{P}(B)$. Since we know $A \subseteq A$, we know $A \in \mathcal{P}(A)$.

Since $\mathcal{P}(A)=\mathcal{P}(B)$, we know that $A \in \mathcal{P}(B)$.

Therefore, $A \subseteq B$ and $A=B$.

Is this proof okay?

Edit: I should note this isn't probability, $\mathcal{P}$ is the power set.
You can prove: $$\bigcup \mathcal{P}(A)=A$$$$\bigcup\mathcal{P}(B)=B$$ and by hypothesis you have $\mathcal{P}(A)=\mathcal{P}(B)$ therefore $$B=\bigcup\mathcal{P}(B)=\bigcup\mathcal{P}(A)=A$$
If we know $A$ we can determine unequivocally $\mathcal{P}(A)$ and if we know $\mathcal{P}(A)$ we can determine unequivocally A, as element of  $\mathcal{P}(A)$ which includes any subset , so if $A=B$ then $\mathcal{P}(A)=\mathcal{P}(B)$ and if $A\neq B$ then $\mathcal{P}(A)\neq\mathcal{P}(B)$ what completes the proof of your theorem.

Another way of proving:
$$(\mathcal{P}(A)=\mathcal{P}(B))\implies [(A\in\mathcal{P}(B)) \wedge (B\in\mathcal{P}(A))]\implies [(A \subseteq B) \wedge (B \subseteq A)]\\
[(A \subseteq B) \wedge (B \subseteq A)] \implies [(A\cup B=B) \wedge (B\cup A=A)]\implies (A=B)$$

PS. -1 as rating of my answer is more against reputation of math.stackexchange.com , than mine.