Find all finite sets $A$ so that $A\times\mathcal P(A) =\mathcal P(A)\times A$, where $\mathcal P(A)$ is the power set of $A$.

Hint 1: $\mathcal P (A)$ always has more elements than $A$. In particular $A \neq \mathcal P(A)$.

Hint 2: If $A \times B =B \times A$ then you can prove that either $A=B$ or one of $A,B$ is empty.

Indeed, if you assume by contradiction that you have $A \neq B$ and that neither $A$ nor $B$ is empty, then you can find an element $x$ which is in exactly one of the sets $A$ or $B$. Now, by using that neither $A$ nor $B$ is empty, you can pick some $y$ in the other set. Look at $(x,y)$.

Hint 3 Conclude that $A=\emptyset$ is the only solution.

Assuming that "x" is the Cartesian product of two sets, we have the definition

For two sets $A$ and $B$, the cartesian product is given by $$A\times B=\{ (a,b):a\in A, b\in B\}$$

Suppose that $A\times P(A)$ for some set $A$. Then, each element of $P(A)$ must be in $A$. However, by Cantor's theorem, we have that $|A|<|P(A)|$ for any $A$, thus there is at least one element of $P(A)$ that is not in $A$.
Thus it cannot be true that $A\times P(A)=P(A)\times A$ for any set $A$.