A Prime $\mathcal P$-filter is contained in a unique $\mathcal P$-ultrafilter? Some backround:
Let $\mathcal P$ be a class of subsets of a topological space such that if $P_1$ and $P_2$ are sets from $\mathcal P$ then $P_1\cap P_2$ and $P_1\cup P_2$ belong to $\mathcal P$. A $\mathcal P$-filter $\mathcal F$  is a collection of nonempty  elements of $\mathcal P$ closed for finite intersections and such that for any $P_1\in \mathcal F$ and $P_1\subseteq P_2\in \mathcal P$ we have $P_2\in \mathcal F$.
A $\mathcal P$-filter $\mathcal F$ is said to be prime if whenever $P_1$ and $P_2$ belong to $\mathcal P$ and $P_1\cup P_2\in \mathcal F$, then $P_1\in \mathcal F$ or $P_2\in \mathcal F$. A $\mathcal P$-ultrafilter is just a maximal $\mathcal P$-filter.
My question is, is every prime $\mathcal P$-filter contained in a unique $\mathcal P$-ultrafilter?; this is exercise 12E.6 of Willard's General Topology.
I have proved that if $\mathcal F$ is a $\mathcal P$-ultrafilter and $P\in \mathcal P$ is such that $P\cap F\neq \emptyset$ for all $F\in \mathcal F$, then $P\in \mathcal F$. I think this must be used in the proof but I don't know how. 
All hints are appreciated.   
 A: In general it's false, I think, even for finite collections. 
The question is really about distributive lattices, as the commenters already said, and there are standard examples of distributive lattices such that a prime filter need no be extendible to a unique ultrafilter, I just have to instantiate
such an example as a collection of subsets of a topological space..:
Let $X = \mathbb{R}$, say, and $\mathcal{P} = \{[0,9],[0,6],[3,9],[3,6],[3,5],[4,6],[4,5]\}$ (the lattice diagram is a "double diamond").
Then $\mathcal{F} = \{[0,9], [3,9]\}$ is a prime filter, but both $\mathcal{U} = \mathcal{P}\setminus \{[3,5],[4,5]\}$ and $\mathcal{U}' = \mathcal{P} \setminus \{[4,6],[4,5]\}$ are ultrafilters extending $\mathcal{F}$. 
A: It's false in general, and even for cases of interest to Williard. 
Let ${\cal P}$ be the collection of open subsets of the real line. Let ${\cal F}$ be the filter of open sets containing 0. This is clearly a prime filter.
Now, every set of the form $(0,\frac{1}{n})$ intersects every element of ${\cal F}$, so there is an open ultrafilter ${\cal F}_1$ containing ${\cal F}$ and also every set of this form. 
But, in the same way, every set of the form $(-\frac{1}{n},0)$ intersects every element of ${\cal F}$, so there is an ultrafilter ${\cal F}_2$ containing ${\cal F}$ and all of these.
Clearly, ${\cal F}_1$ and ${\cal F}_2$ are different and ${\cal F}$ is contained in both.
