BPI is equivalent to arbitrary-set propositional compactness

Assume BPI and finite satisfiability of $T$. The set $2^P$ is nonempty, with the constant-zero valuation as an explicit member. The subsets whose membership depends on finitely many coordinates form a Boolean algebra $A$: a union or intersection of two such sets depends on the union of their finite supports, and complement has the same support. The truth set $\mathrm{⟦}\varphi \mathrm{⟧}$ of any finite formula is in $A$, by recursion on its connectives. Finite satisfiability says every finite intersection of the truth sets of members of $T$ is nonempty. Their finite intersections and upward closure therefore form a proper Boolean filter in $A$, including $2^P$ from the empty intersection.

Separately, assume propositional compactness and take a nontrivial Boolean algebra $B$. Use letters $p_b$ for $b\in B$ and the theory consisting of $p_1$, $\mathrm{\neg }{p}_0$, and all equations $p_{\mathrm{\neg }b}\leftrightarrow \mathrm{\neg }{p}_b$, $p_{b\wedge c}\leftrightarrow (p_b\wedge p_c)$ and $p_{b\vee c}\leftrightarrow (p_b\vee p_c)$. Any finite fragment names finitely many Boolean elements. As in finite distributive expansion, the joins of all cells obtained by choosing each named element or its complement form a finite subalgebra containing them. It is nontrivial because it contains distinct $0,1$. By F2 an atom gives a character on that subalgebra, satisfying every equation in the fragment. Extend the letter valuation by zero outside that subalgebra. Thus the theory is finitely satisfiable.

Under BPI, F1 extends the filter of step 1.1 to an ultrafilter $U$. For $p\in P$, set $v(p)=1$ precisely when $\{w:w(p)=1\}\in U$. An ultrafilter decides complements: adjoining a missing element makes a filter improper, so an old element is disjoint from it, forcing its complement into the old filter. Together with finite meet closure, this gives $a\wedge b\in U$ exactly when both $a,b\in U$, and $a\vee b\in U$ exactly when at least one belongs to $U$. Recursion on formulas now proves $v(\varphi )=1$ exactly when $\mathrm{⟦}\varphi \mathrm{⟧}\in U$, checking negation, conjunction and disjunction by these identities and both truth constants by properness. All truth sets from $T$ lie in $U$, so $v$ satisfies $T$.

A model of this propositional theory yields $h:B\to 2$, $h(b)=v(p_b)$, preserving all Boolean operations and both bounds by the equations. Its zero fibre is a proper ideal; if $h(b\wedge c)=0$, the two values in $2$ cannot both be $1$, so one of $b,c$ lies in that ideal. Hence it is prime, proving BPI. Empty $P$ in the forward direction causes no difficulty: $2^P$ is a singleton, and finite satisfiability excludes the false truth set. Empty $T$ is satisfied by the constant-zero valuation. No infinite choice was used in the finite fragment argument. QED.