Answer
See solution and explanation below.
Work Step by Step
Let $A$ and $B$ be sets. We want to show $\mathcal{P}(A) \subseteq \mathcal{P}(B)$ if and only if $A \subseteq B$. So we will show $\mathcal{P}(A) \subseteq \mathcal{P}(B) \implies A \subseteq B$ and $A \subseteq B \implies \mathcal{P}(A) \subseteq \mathcal{P}(B)$.
First, suppose $\mathcal{P}(A) \subseteq \mathcal{P}(B)$. To show $A \subseteq B$, we will show $x \in A \implies x \in B$ for all $x \in A$. So let $x \in A$, then $\{x\} \subseteq A$. This means $\{x\} \in \mathcal{P}(A)$. Thus since $\mathcal{P}(A) \subseteq \mathcal{P}(B)$, $\{x\} \in \mathcal{P}(B)$, which means $\{x\} \subseteq B$, and, therefore, $x \in B$. Hence $A \subseteq B$.
For the converse, suppose $A \subseteq B$. We must show $\mathcal{P}(A) \subseteq \mathcal{P}(B)$. So let $C \in \mathcal{P}(A)$. Then $C \subseteq A$. Since $A \subseteq B$, every subset of $A$ is a subset of $B$. Hence $C \subseteq B$. But this means $C \in \mathcal{P}(B)$. Therefore, $\mathcal{P}(A) \subseteq \mathcal{P}(B)$, and, thus, $\mathcal{P}(A) \subseteq \mathcal{P}(B)$ if and only if $A \subseteq B$.
