PFA implies the pseudointersection number exceeds omega-one

Let $P$ consist of pairs $(s,F)$ with $s\in [\omega {]}^{<\omega }$ and $F\in [\mathcal{A}{]}^{<\omega }$. Put $(t,G)\le (s,F)$ exactly when $s\subseteq t$, $F\subseteq G$, and $t\setminus s\subseteq \bigcap F$, taking $\bigcap \mathrm{\varnothing }=\omega$. For a fixed finite stem $s$, every finite collection of conditions with that stem has the common extension whose side set is the union of their side sets. Since there are countably many finite subsets of $\omega$, $P$ is sigma-centered and therefore ccc.

For each $a\in \mathcal{A}$, the set $E_a=\{(s,F):a\in F\}$ is dense, because adding $a$ to $F$ changes no stem. For each $n<\omega$, let $D_n=\{(s,F):(\mathrm{\exists }k\in s)k\ge n\}$. Given $(s,F)$ outside $D_n$, the strong finite intersection property makes $\bigcap F$ infinite, so choose $k\in \bigcap F$ with $k\ge n$ and extend the stem by $k$; hence $D_n$ is dense. The family of all $E_a$ and $D_n$ has cardinality at most ${\omega }_1$.

By F2 and F3, choose a filter $G\subseteq P$ meeting every set from step 2.1, and put $b=\bigcup \{s:(s,F)\in G\}$. Meeting all $D_n$ makes $b$ unbounded in $\omega$, hence infinite. Fix $a\in \mathcal{A}$ and choose $(s,F)\in G\cap E_a$. For any $(t,H)\in G$, directedness gives $(u,K)\in G$ below both; the order relative to $(s,F)$ gives $u\setminus s\subseteq \bigcap F\subseteq a$, and $t\subseteq u$. Thus $t\setminus a\subseteq s$, and after taking the union, $b\setminus a\subseteq s$ is finite. Therefore $b{\subseteq }^{\ast }a$ for every $a\in \mathcal{A}$, so $b$ is an infinite pseudointersection.

Since every at-most-${\omega }_1$ strong-finite-intersection family has such a pseudointersection, no family witnessing the definition of $\mathfrak{p}$ has cardinality at most ${\omega }_1$. By F1, $\mathfrak{p}>{\omega }_1$. Empty and finite $\mathcal{A}$ are included: the same forcing works, and for $\mathcal{A}=\mathrm{\varnothing }$ the constructed $b$ is simply infinite.