= Solution
Here the Hom groups must be taken over $A$; the printed subscript $B$ in this part has no defined coefficient ring. For an arbitrary <module> $P$, form the <free module>
$$
F=\bigoplus_{p\in P}Ae_p,
$$
and define a <module homomorphism> $\pi:F\to P$ by $e_p\mapsto p$. Each element of $F$ has finite support, so this prescription extends uniquely and is well-defined. It is <surjective> because every $p\in P$ is the image of $e_p$. Thus every module is a quotient of a <free module>.
Let $K=\ker\pi$. The <short exact sequence> $0\to K\to F\xrightarrow{\pi}P\to0$ and the defining exactness condition for a <projective module> make
$$
\operatorname{Hom}_A(P,F)\longrightarrow\operatorname{Hom}_A(P,P),\qquad s\longmapsto\pi s
$$
<surjective>. Lift $1_P$ to a map $s:P\to F$. Then $\pi s=1_P$, so $s$ is <injective>. Every $f\in F$ has the decomposition
$$
f=(f-s\pi(f))+s\pi(f),
$$
whose first term lies in $K$ and second lies in $s(P)$. If $s(p)\in K$, applying $\pi$ gives $p=0$, so the intersection is zero. Consequently
$$
\boxed{F=K\oplus s(P)\cong K\oplus P.}
$$
This is a <split short exact sequence> and proves that <projective modules are direct summands of free modules>, including modules that are not finitely generated.
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