Isomorphism of $V_\lambda$ and the ultraproduct of $V_{\lambda_{\mathrm{otp}(x)}}, x \in \mathcal{P}_\kappa({\lambda})$ by a normal fine measure. This question is actually exercise (20.5) from Set Theory by Thomas Jech. The original statement is:

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

Here, $\lambda_x$ is the order type of $x$.
I was able to shown (if I'm not mistaken) that if $M$ is the transitive collapse of $\mathrm{Ult}_U(V)$, and $N$ is the transitive collapse of $\mathrm{Ult}_U\{V_{\lambda_x}: \ x \in \mathcal{P}_\kappa({\lambda})\}$, then $N = V_\lambda\cap M$.
I have done this using a rank argument: Let $a \in M$ such that $\mathrm{rk}(a)<\lambda$, and let $f$ be the function on $\mathcal{P}_\kappa(\lambda)$ that represents $a$ in $\mathrm{Ult}_U(V)$ (i.e. $\pi ([f]_U) = a$ where $\pi$ is the collapsing function). Then as $x \mapsto \lambda_x$ represents $\lambda$, we have $\mathrm{rk}(f(x)) < \lambda_x$  for almost all $x$. So we may assume that this holds for all $x$, and then $[f]_U \in \mathrm{Ult}_U\{V_{\lambda_x}: \ x \in \mathcal{P}_\kappa({\lambda})\}$.
But why would $V_\lambda^M = V_\lambda$ hold in general? Moreover, what about the fact that $\left|N\right| \le \kappa^{\lambda^{<\kappa}}$, couldn't $V_\lambda$ be much larger?
 A: As pointed out in a comment by spaceisdarkgreen, an almost identical question has been posted on MO in the past:
https://mathoverflow.net/q/44289
It has recived an answer by Joel David Hamkins:
https://mathoverflow.net/a/44291
For completeness, here is the answer given by Hamkins:

In general, $\lambda$-supercompactness, if consistent, does not imply $\lambda$-strongness. One can see this by observing that the
  smallest cardinal $\kappa$ that is $\kappa^+$-supercompact is never
  $\kappa^+$-strong, and in fact, cannot be even $(\kappa+3)$-strong.
  The reason is that $\kappa^+$-supercompactness is witnessed by a
  measure on $P_\kappa\kappa^+$, which amounts essentially to (is coded
  by) a subset of $P(\kappa^+)$, and hence is witnessed inside
  $V_{\kappa+3}$. Thus, if $j:V\to M$ were any embedding with critical
  point $\kappa$, by minimality it follows that $\kappa$ is not
  $\kappa^+$-supercompact in $M$, and hence $M$ cannot have the true
  $V_{\kappa+3}$. Thus, $\kappa$ is not $(\kappa+3)$-strong and thus
  definitely not $\kappa^+$-strong.
I haven't looked at the context of the exercise, but perhaps he is
  merely asking you to make the observation that you did in fact make,
  that the ultrapower will give you $V_\lambda^M$? 
For some kinds of $\lambda$, it does follow that
  $\lambda$-supercompactness implies $\lambda$-strongness. For example,
  if $\lambda$ is a beth-fixed point, then every
  $\lambda$-supercompactness embedding is also $\lambda$-strong and even
  $(\lambda+1)$-strong, since in this case $|V_\lambda|=\lambda$. But if
  $\lambda$ is not a beth-fixed point, then a version of my argument
  above will still apply: if $\lambda$ is not a beth-fixed point and
  $j:V\to M$ is a Mitchell minimal $\lambda$-supercompactness embedding
  for $\kappa$, then $\kappa$ is not $\lambda$-supercompact in $M$, and
  this is witnessed inside $V_\lambda$ by the assumption on $\lambda$,
  and so $j$ cannot be a $\lambda$-strongness embedding.

