A Prime $\mathcal P$-filter is contained in a unique $\mathcal P$-ultrafilter?

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 $\mathcal{F}$.

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.