Does $\mathfrak{P}(A)\cong\mathfrak{P}(B)$ imply $A\cong B$?

Question

Does $\mathfrak{P}(A)\cong\mathfrak{P}(B)\implies A\cong B$ hold for arbitrary sets $A, B$?

Notation:

• We write $\mathfrak{P}(M)$ to denote the power set of a set $M$.

• $M_1\cong M_2$ is just an abbreviation for "there is a bijection from the set $M_1$ in the set $M_2$".