The ranks are the same. Since $M$ is a finitely generated $(S/\mathfrak{p})$-module you can write down a finite presentation of $M$ by twisted (in the sense of twisting the grading) copies of $S/\mathfrak{p}$:

$$ \oplus_j (S/\mathfrak{p})(-b_j) \stackrel{\Phi}{\longrightarrow} \oplus_{i=1}^k (S/\mathfrak{p})(-a_i) \longrightarrow M \longrightarrow 0.$$

Passing to $\operatorname{Proj} S/\mathfrak{p}$ this becomes a presentation of $\mathcal{F}_{P}$ with the terms $(S/\mathfrak{p})(-a_i)$ (respectively $-b_j$) replaced by copies of the structure sheaf twisted by $-a_i$ (respectively $-b_j$). Thus the rank of $\mathcal{F}_P$ at a generic point of $\operatorname{Proj} S/\mathfrak{p}$ is $k$ minus the rank of the map $\Phi$ at a generic point of the $\operatorname{Proj}$.

On the other hand, passing to $\operatorname{Spec} S/\mathfrak{p}$, the presentation above becomes a presentation of $\mathcal{F}_A$ with the terms $(S/\mathfrak{p})(-a_i)$ (respectively $-b_j$) replaced by the structure sheaf. The rank of $\mathcal{F}_P$ at a generic point of $\operatorname{Proj} S/\mathfrak{p}$ is again $k$ minus the rank of the map $\Phi$ at a generic point of the $\operatorname{Spec}$.

The scheme $\operatorname{Spec} S/\mathfrak{p}$ is the affine cone over $\operatorname{Proj} S/\mathfrak{p}$, and any point on the $\operatorname{Proj}$ gives rise to a line on the $\operatorname{Spec}$. The rank of $\Phi$ at any point of the $\operatorname{Proj}$ is equal to the rank of $\Phi$ at every point of the corresponding line in the $\operatorname{Spec}$ (except possibly the origin). Thus the rank of $\Phi$ at a generic point of the $\operatorname{Proj}$ is the same as the rank of $\Phi$ at a generic point of the $\operatorname{Spec}$, and therefore the ranks of $\mathcal{F}_P$ and $\mathcal{F}_A$ at the respective generic points are also the same.