For a not empty set $A$ let $\mathcal{P}(A)$ be the power set of $A$. Show that $A$ is not equivalent to $\mathcal{P}(A)$

Question

For a not empty set $A$ let $\mathcal{P}(A)$ be the power set of $A$. Show that $A$ is not equivalent to $\mathcal{P}(A)$

Well I´m trying to show it by contradiccion. So suppouse that $\sigma :A \rightarrow \mathcal{P}(A)$ is surjetive.So then I just need to find a subset $B=\{x: x \in A \quad \textrm{and} \quad x \notin \mathcal{P}\}$. Any suggestions?