Suppose $A \sim B$. Then, there is a one-to-one and onto function $f : A \rightarrow B$. Since $f$ is one-to-one and onto, $f^{- 1} : B \rightarrow A$ exists. Let $g : \wp(A) \rightarrow \wp(B)$ be defined as $g (A) = \{ f (a) |a \in A \}$ and $h : \wp(B) \rightarrow \wp(A)$ as $h (B) = \{ f^{- 1} (b) |b \in B \}$. To prove $g$ is one-to-one, let $C \subseteq A$ and $D \subseteq A$. Suppose $g (C) = g (D)$. Then, $h (g (C)) = \{ f^{- 1} (f (c)) |c \in C \} = C = D = \{ f^{_{- 1}} (f (d)) |d \in D \} = h (g (D))$. To prove $g$ is onto, let $C \subseteq B$. Then, $h (C) \subseteq A$, and $g (h (C)) = \{ f (f^{- 1} (c)) |c \in C \} = C$.
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Your proof is correct but can be improved
What you could improve: define $g$ for a set $X \in \wp(A)$, not just for $A$ (although it is clear what you mean) and the same for $h$. Furthemore, you could improve the part where $g$ is one-to-one by instead of writing $$h(g(C))=\{f^{−1}(f(c))|c\in C\}=C=D=\{f^{−1}(f(d))|d\in D\}=h(g(D))$$ write $$C=\{f^{−1}(f(c))|c\in C\}=h(g(C))=h(g(D))=\{f^{−1}(f(d))|d\in D\}=D$$ (as the reader might think that you have assumed $C=D$ in the first part and proved $h(g(C)) = h(g(D))$ instead of the opposite.
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