For every set $x$ there is exactly one set whose elements are precisely the subsets of $x$

Power Set supplies a set $y$ that contains every $z$ all of whose elements belong to $x$; by the meaning of inclusion, $y$ contains every $z$ with $z\subseteq x$.

Apply Separation to $y$ with the formula $\phi (z,x):=\mathrm{\forall }t(t\in z\to t\in x)$ and the parameter $x$: the set $p:=\{z\in y:z\subseteq x\}$ exists, and for every $z$, $z\in p$ holds if and only if $z\in y$ and $z\subseteq x$.

If $z\subseteq x$ then $z\in y$ by step 1.1, so the condition "$z\in y$ and $z\subseteq x$" reduces to $z\subseteq x$; hence $z\in p$ holds if and only if $z\subseteq x$, which proves existence.

If $p^{\mathrm{\prime }}$ also satisfies "$z\in p^{\mathrm{\prime }}$ if and only if $z\subseteq x$" for every $z$, then $z\in p$ holds if and only if $z\in p^{\mathrm{\prime }}$ for every $z$, so $p=p^{\mathrm{\prime }}$; existence and uniqueness together give the statement.