= Solution
Write $[f]$ for the equivalence class of $f:\kappa\to V_\lambda$ in the <ultrapower>
$$
N=(V_\lambda)^\kappa/U,
$$
and define its membership relation by
$$
[g]\mathrel E[f]
\quad\Longleftrightarrow\quad
\{\xi<\kappa:g(\xi)\in f(\xi)\}\in U.
$$
The <kappa-complete filter> property makes $E$ <well-founded relation>[well-founded]: an infinite descending $E$-chain would give countably many members of $U$ whose intersection belongs to $U$, and every index in that intersection would yield an infinite descending membership chain, contradicting the <Axiom of foundation>. The relation is <extensional relation>[extensional] by <Łoś theorem>[Łoś's theorem].
The <Mostowski collapse theorem> therefore gives a unique isomorphism $\pi:(N,E)\to(M,\in)$ onto a <transitive set> $M$. Recursively, the notation missing from the printed formula may be defined by
$$
\pi([f])=\{\pi([g]):[g]\mathrel E[f]\}.
$$
The value is independent of the representative because it is defined on the ultrapower class $[f]$. Moreover $M\subseteq V_\lambda$: every $f:\kappa\to V_\lambda$ has its range contained in some $V_\alpha$ with $\alpha<\lambda$, since $\kappa<\lambda$ and the <strongly inaccessible cardinal> $\lambda$ is <regular cardinal>[regular]; induction on the resulting rank bound keeps $\pi([f])$ inside $V_\lambda$.
Define the <ultrapower embedding>
$$
j:V_\lambda\longrightarrow M,
\qquad
j(x)=\pi([\operatorname{const}_x]).
$$
The constant-function map into $N$ is <elementary embedding>[elementary] by Łoś's theorem, and $\pi$ is an isomorphism, so their composite $j$ is elementary.
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