The Mitchell order Here, on page 358, I would like to know
if the $M$-ultrafilter $U_i$ on $κ$ is iterable for all $i$'s? 
How do we use and prove this fact?       
And how can I see that $P^M(κ)=P(κ)$ and $M^κ⊂M$?
And finally how can I see that $\{U_i\}_{i=1}^{\infty}$ is in $M$?
 A: First, let's show that $M$ is closed under sequences of length $<\kappa$ - this will address your last question. Let's begin by remembering how sequences transform under $j$ in general:

Suppose $f$ is a function with domain $\alpha$. Then for $\beta<\alpha$ we have $j(f)(j(\beta))=j(f(\beta))$.

This leaves room for several changes in moving from $f$ to $j(f)$. Obviously we can have $j(f(\beta))\not=f(\beta)$, but more importantly here the domain may change: we could have $j(\alpha)>\alpha$, in which case we will have inputs to $j(f)$ not corresponding to any inputs to $f$, and moreover the bits of $j(f)$ corresponding to bits of $f$ will move (if $j(\beta)>\beta$, then $j(f(\beta))$ is the $j(\beta)$th term of $j(f)$, not the $\beta$th term). The point here is that if $\alpha$ is "small," then $f$ can't change too much when we hit it with $j$.

First, let's show that $M$ is closed under $(<\kappa)$-sequences. (This will show why $\langle U_i\rangle_{i<\omega}$ is in $M$, since each $U_i$ is in $M$ and $\omega<\kappa$.) Suppose $a_\eta\in M$ with $a_\eta=j(b_\eta)$, for $\eta<\theta$ for some fixed $\theta<\kappa$. Then we have $j(\langle b_\eta\rangle_{\eta<\theta})=\langle a_\eta\rangle_{\eta<\theta}$: this is because $\theta<crit(j)$, so doesn't get "stretched." Phrased another way, any function $f$ with domain $\theta$ satisfies $$j(f)=\{(a, b): a\in\theta, b=j(f(a))\}.$$
Now what about sequences of length $\kappa$? Here we have a problem, since the sequence will "stretch" - e.g. $j(\langle \eta\rangle_{\eta<\kappa})=\langle\eta\rangle_{\eta<j(\kappa)}$. However, this is still fixable: all the "nonstandard" terms of the sequence that $j$ adds come after all the "standard" terms. If $f$ is a function with domain $\kappa$, we have $$j(f)\upharpoonright\kappa=\{(a,b): a\in\kappa, b=j(f(a))\}.$$
Now elements of $\mathcal{P}(\kappa)$ are (basically equivalent to) $\kappa$-sequences of ordinals, and every ordinal is in $M$ - so $\mathcal{P}(\kappa)\subseteq M$, hence $\mathcal{P}(\kappa)=\mathcal{P}(\kappa)^M$.

As to iterability, I don't see that it plays any role here, but I could be reading too quickly.
A: Here's an alternative proof of the fact that $\mathcal{P}^M(\kappa) = \mathcal{P}(\kappa)$:
Since $M \subseteq V$ is transitive, we have $\mathcal{P}^M(\kappa) \subseteq \mathcal{P}(\kappa)$. (Note that "$x \subseteq \kappa$" is expressible via a $\Sigma_0$-formula and hence absolute between transitive structures. Hence $M$ correctly identifies whether a given element $x \in M$ is a subset of $\kappa$ or not - but it could miss subsets of $\kappa$.)
Let $x \subseteq \kappa$. Let $\pi \colon V \to M$ be the canonical elementary embedding. Then 
$$M \models \pi(x) \subseteq \pi(\kappa)$$
and since this is absolute (see above), we have $\pi(x) \subseteq \pi(\kappa)$. Now, for any $\xi < \kappa$
$$
\begin{align*}
\xi \in x & \iff V \models \xi \in x \\
& \iff M \models \underbrace{\pi(\xi)}_{= \xi} \in \pi(x) \\
& \iff \xi \in \pi(x)
\end{align*}
$$
and thus $x = \pi(x) \cap \kappa \in M$. (Here we use that $M$ computes $a \cap b$ correctly for any $a,b \in M$.)
Essentially the same argument actually shows
Claim. $V_{\kappa +1}^M = V_{\kappa +1}$.
Proof. By the same argument as above (and absoluteness of the rank function) we have $V_{\kappa +1}^M \subseteq V_{\kappa +1}$. Conversely let $x \in V_{\kappa +1} = \mathcal P(V_{\kappa})$. It suffices to show that
$$
x = \{ y \in \pi(x) \mid \mathrm{rank}_{\in}(y) < \kappa \} = \{ y \in \pi(x) \mid M \models \mathrm{rank}_{\in}(y) < \kappa \}.
$$
Since "$\mathrm{rank}_{\in}(y) = \alpha$" is absolute between transitive models of $\mathrm{ZFC}$, we get the equality of the sets on the right hand side. Furthermore for any $y \in x$ we have $\mathrm{rank}_{\in}(y) < \kappa$ and thus $y = \pi(y) \in \mathrm{RHS}$. (Recall that $\mathrm{crit}(\pi)$ is the minimal rank of an element $z$ such that $\pi(z) \neq z$.) Conversely, if $y \in \mathrm{RHS}$, then $y = \pi(y) \in \pi(x)$ and thus - by elementarity - $y \in x$. Q.E.D.

As promised here, I'll add a proof that $M$ is in fact closed under $\kappa$-sequences. Let $(x_i \mid i < \kappa)$ be a sequence such that $x_i \in M$ for all $i < \kappa$. Fix, for each $i < \kappa$, a function
$$
f_i \colon \kappa \to V
$$
such that $x_i = [f_i]$. (This is possible because $M= \mathrm{Ult}_{U_0}(V)$.) Define a function
$$
g \colon \kappa \to V
$$
as follows: For each $\xi < \kappa$
$$
g(\xi) \colon \xi \to V
$$
is such that for all $i < \xi$
$$
g(\xi)(i) = f_{i}(\xi).
$$
Using that $\kappa = [\mathrm{id}]$ (which is equivalent to $U_0$ being normal) and Łoś's Theorem, you can verify that
$$
[g] \colon \kappa \to M
$$
(in $M$) and that for all $i < \kappa$
$$
\{ \xi < \kappa \mid g(\xi)(i) = f_{i}(\xi) \} = \kappa \setminus (i+1) \in U_0,
$$
so that by Łoś's Theorem, for all $i < \kappa$,
$$
[g](i) = [f_i]
$$
and thus
$$
(x_i \mid i < \kappa) = ([f_i] \mid i < \kappa) = [g] \in M.
$$
