= Solution
Use the embedding characterization of a <supercompact cardinal>: for every cardinal $\mu\geq\kappa$ there is an <elementary embedding> $j:V\to M$ into a <transitive class> with critical point $\kappa$, $j(\kappa)>\mu$ and $M^\mu\subseteq M$. In the <Lévy hierarchy>, a $\Sigma_2$ sentence can be put into the form $\exists x\,\forall y\,\theta(x,y)$ with bounded matrix $\theta$. The same argument allows parameters $p\in V_\kappa$, which the embedding fixes.
Suppose $V\models\exists x\,\forall y\,\theta(x,y,p)$ and choose a witness $a$. Pick an <ordinal> $\lambda$ with $a,p\in V_\lambda$, and a cardinal $\mu\geq\kappa$ large enough that $\mu\geq|V_\lambda|$ and $\mu>\lambda$. Take the corresponding supercompact embedding. Its closure implies $V_\lambda\subseteq M$: induct on rank, enumerate each set in this rank segment by at most $\mu$ elements already in $M$, and use $M^\mu\subseteq M$ followed by taking the range. Thus $a\in M$ and $\operatorname{rank}(a)<j(\kappa)$.
Bounded formulas are absolute between <transitive classes>. Since $\forall y\,\theta(a,y,p)$ holds in $V$, it holds in $M$ and then in the transitive rank segment $V_{j(\kappa)}^M$. This segment contains $a,p$, so
$$
M\models V_{j(\kappa)}\models\exists x\,\forall y\,\theta(x,y,p).
$$
Apply elementarity to the first-order assertion that this sentence holds in the indicated <cumulative hierarchy> level, and use $j(p)=p$. We obtain the <supercompact Sigma-two downward reflection> conclusion
$$
\boxed{V\models\sigma(p)\Longrightarrow V_\kappa\models\sigma(p)\qquad(\sigma\in\Sigma_2,\ p\in V_\kappa).}
$$
The closure requirement is used to put the original witness into the target; elementarity alone would not do that.
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