Ultrafilters on finite boolean algebras I am asked to prove a special case of Stone duality, namely that $B\cong \mathcal{P}(\text{Ult}(B))$ by the map $\phi:B\to \mathcal{P}(\text{Ult}(B))$ given by the homomorphism
$$
\phi(x)=\{V\in \text{Ult}(B) \;|\; x\in V\},
$$
where $B$ is a finite boolean algebra and $\text{Ult}(B)$ denotes the set of ultrafilters contained in $B.$ I have shown that $\phi$ is injective. Now I need that $\phi$ is surjective. I have considered two ways to do this.
The first way is to simply take a collection of ultrafilters $\mathcal{U}=\{U_1,\ldots U_n\}$ and find some $b\in B$ with $\phi(b)=\mathcal{U}.$ If $U=U_1\cap\cdots\cap U_n$ and $b=\bigwedge_{x\in U}x,$ then it seems like $\phi(b)$ should be $\mathcal{U}.$ This amounts to showing that if $V$ is an ultrafilter and $V\supseteq U,$ then $V=U_k$ for some $k.$ However, I do not know how to show this.
The second way is to simply compute the cardinalities of $B$ and $\mathcal{P}(\text{Ult}(B)).$ This requires knowing that $|B|=2^m$ for some $m$ and $|\text{Ult}(B)|=m.$ Again, I do now know how to show this.
Maybe I'm missing something rather simple, but I'm stuck. Any help is appreciated.

Here's an easier way to show surjectivity: since $\mathcal{P}(\text{Ult}(B))$ is generated by singletons, it suffices to show that every singleton is in the image of $\phi$.  In other words, given an ultrafilter $U$ on $B$, we want to find $x\in B$ which is in $U$ and no other ultrafilters.
I encourage you to try to finish the proof from here on your own.  One way to do so is hidden below.

 For each ultrafilter $V\neq U$, choose $x_V\in U\setminus V$.  Now let $x$ be the meet of all of these elements $x_V$.

