In Chapter 14 of Jech's Set Theory, Theorem 14.24 states the following (see here for the definition of $V^B$):
Theorem 14.24. Every axiom of ZFC is valid in $V^B$.
For the proof of Power Set, he writes the following:
We prove that for every $X\in V^B$ there is $Y\in V^B$ such that $$(14.24)\qquad ||\forall u(u\subseteq X\to u\in Y)||=1.$$ (cf (1.9)). Here we let $$\text{dom}(Y)=\{u\in V^B:\text{dom}(u)=\text{dom}(X)\text{ and } u(t)\leq X(t)\text{ for all } t\},\\ Y(t)=1\text{ for all }t\in \text{dom}(Y).$$ To verify that $Y$ satisfies (14.24) we use the following observation: if $u\in V^B$ is arbitrary, let $u'\in V^B$ be such that $\text{dom}(u')=\text{dom}(X)$ and $u'(t)=X(t)\cdot ||t\in u||$ for all $t\in \text{dom}(X)$. Then $$||u\subseteq X||\leq ||u=u'||$$ which makes it possible to include in $\text{dom}(Y)$ only the "representative" $u$'s.
I am currently stuck trying to understand why $||u\subseteq X||\leq ||u=u'||$. I understand how to complete the proof from here, as $||u=u'||\leq Y(u')\cdot ||u=u'||\leq ||u\in Y||$ since $u'\in Y$, but getting here is proving difficult for me. I have written out $$||u\subseteq x||=\prod_{t\in \text{dom}(u)}(u(t)\Rightarrow||t\in X||)\\\ ||u=u'||=\left(\prod_{t\in \text{dom}(u)}(u(t)\Rightarrow||t\in u'||)\right)\cdot \left(\prod_{t\in \text{dom}(u')}(u'(t)\Rightarrow||t\in u||)\right)$$ and can see that the latter factor of the second term is $1$ since $u'(t)\leq ||t\in u||$ for all $t$, so it would suffice to show that $||t\in X||\leq ||t\in u'||$ for all $t\in \text{dom}(u)$, for which it would suffice to show $||X(s)||\cdot ||s=t||\leq X(s)\cdot ||s=t||\cdot ||t\in u||$ for all $s\in \text{dom}(X)$. However, I see no reason for this to be true and as such am stuck here.
Think about it heuristically: We have essentially defined $u':=X\cap u$ and you're essentially trying to prove $t\in X\to t\in u',$ which is hopeless... unless of course you already know that $t\in u,$ which you do.
So this motivates a rewrite $$ \Vert u\subseteq X \Vert= \prod_t\left(u(t)\Rightarrow \Vert t\in X\Vert\right)=\prod_t\left(u(t)\Rightarrow \sum_s \Vert t=s\Vert \cdot X(s)\right) \\=\prod_t \sum_s u(t)\Rightarrow\Vert t=s\Vert\cdot X(s)$$ and similarly for the other expression. And then we can see that what we really want to show is that $$ \left(u(t)\Rightarrow\Vert t=s\Vert\cdot X(s)\right) \le\left( u(t)\Rightarrow\Vert t=s\Vert\cdot X(s)\cdot \Vert t\in u\Vert\right).$$ But you can make short work of this using the fact that $u(t)\le \Vert t\in u\Vert.$