This is exercise 20.5 out of Jech:

Let $\lambda \geq \kappa$ and let $U$ be a normal measure on $P_{\kappa}(\lambda)$. The ultraproduct $\mathrm{Ult} _U \{ (V _{\lambda _x},\in) : x \in P _{\kappa}(\lambda) \}$ is isomorphic to $(V _{\lambda}, \in)$

Here $\lambda _x$ simply denotes the order type of $x$. The function $x \mapsto \lambda _x$ represents $\lambda$ in the ultrapower of $V$ by $U$. Unless I'm mistaken, the ultraproduct mentioned will be $V^M _{\lambda} = V _{\lambda} \cap M$ where $M$ is the ultrapower of $V$ by $U$. I don't see why this would be $V _{\lambda}$ itself, since I don't see why $V _{\lambda} \subset M$. Clearly $H _{\lambda ^+} \subset M$ since $M$ is closed under $\lambda$ sequences, but I sort of doubt that $V _{\lambda} \subset M$ -- I figure if $\lambda$-supercompactness implied $\lambda$-strongness, I would've seen that mentioned somewhere.

So did I make a mistake in computing the ultraproduct, or is there a mistake in the exercise, or does $\lambda$-supercompactness imply $\lambda$-strongness for some reason I'm not seeing?