For brevity, I omitted the proof of the claim. The [] notation indicates the basic open set in the stone space over M that the formula determines. The bar notation indicates a finite sequence of variables or elements. The red arrow indicates the implication that I suspect Marker used in the step.
However, I am not sure how it follows. I agree that the number of types containing phi(v) is greater than or equal to aleph_1. But not all of these types need to be realized in M, so why should the number of elements in x in M that cause phi(x) to be true be greater than or equal to aleph_1?
Moreover, why does it say "by choice of p"? I don't see why the choice of p is useful for that particular step.
Any help would be appreciated.
I am not sure if this answer is the argument that Marker had in mind, but it works anyway. If you replace $[ \phi(v) ]$ throughout by $\{x \in M : \phi(x)\}$ (and similarly for other uses of $[ \; ]$, like $[\phi(v) \land \psi(v)]$), then the entire argument (including the claim) still works. So instead of working with opens in the Stone space, you simply work directly with the sets of realisations in $M$.
The Stone space itself is used nowhere, so I suspect that this may actually be the 'right' argument. At least it is more transparent.
The bit "By choice of $p$" refers to the fact that $|\{x \in M : \phi(x) \land \neg(\delta_0(x) \land \ldots \land \delta_n(x))\}| \leq \aleph_0$.