I must show that the following properties for an $A$-module $P$ are equivalent:
1) The functor $Hom(P,-)$ is exact.
2) There is an $A$-module $Q$ such that $P \oplus Q$ is free.
3) Every short exact sequence of $A$-modules of the form $0 \rightarrow N \rightarrow M \rightarrow P \rightarrow 0$ splits.
4) For every epimorphism $p: M \rightarrow Q$ of $A$-modules and every homomorphism $f:P \rightarrow Q$, there is a homomorphism $g: P \rightarrow M$ such that $f = p \circ g$.
I could go this far:
$(2) \Rightarrow (4)$
Let $Q$ be a module such that $P \oplus Q$ is a free module and let $B = {b_i}_{i \in I}$ be a basis of $P \oplus Q$. Since $g$ is an epimorphism for every $i \in I$, we can find $m_i \in M$ such that $g(m_i) = f(b_i)$. Define $\overline{h}: P \oplus Q \rightarrow M$ by $\overline{h}(\sum_i r_ib_i) := \sum_i r_im_i$. Since $B$ is a basis of $P \oplus Q$, the map is a well-defined $A$-module homomorphism and $g \circ \overline{h} = f$, as desired.
$(4) \Rightarrow (3)$
Since $g$ is an epimorphism, there is $h: P \rightarrow M$ such that $g\circ h = id_P$. Therefore, by Theorem 3.6.5, the sequence splits.
$(3) \Rightarrow (2)$
Consider the canonical epimorphism of $A$-modules $f: \bigoplus_{p \in P} R \rightarrow P$. This gives a short exact sequence $0 \rightarrow Ker(f) \rightarrow \bigoplus_{p\ in P} R \rightarrow P \rightarrow 0$. By assumption on P this sequence splits, so we obtain $P \oplus Ker(f) \simeq \bigoplus_{p \in P} R$ and, thus, P is projective.