= Solution
Fix $x\in V_\delta$. Since $\delta$ is a <limit ordinal>, choose $\beta<\delta$ with $x\in V_\beta$. Transitivity gives $x\subseteq V_\beta$. Every subset $u\subseteq x$ therefore belongs to $V_{\beta+1}=\mathcal P(V_\beta)$, and consequently
$$
\mathcal P(x)\subseteq V_{\beta+1},\qquad\mathcal P(x)\in V_{\beta+2}\subseteq V_\delta.
$$
The last inclusion uses $\beta+2<\delta$, since a finite number of successor steps remains below a <limit ordinal>. In particular every subset of $x$ is also an element of the <transitive set> $V_\delta$.
Take $y=\mathcal P(x)$. For any $u\in V_\delta$, transitivity ensures that the internal statement $u\subseteq x$ agrees with the external one. Hence
$$
(V_\delta,\in)\models\forall u\,(u\in y\leftrightarrow u\subseteq x).
$$
Such a $y\in V_\delta$ exists for every $x\in V_\delta$, so \b[$(V_\delta,\in)$ satisfies the <Axiom of power set>]. This <power-set axiom at limit ranks> actually holds for every nonzero <limit ordinal>; the stated restriction $\delta>\omega$ is sufficient.
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