The following statements are equivalent
If $P$ is an $R-$module, and if it sastisfies the following equivalent properties then it is called a projective module.
$1$- Every exact sequence of the form $0 \to A \to B \to P \to 0$ splits.
$2$-There exists a $R-$module $Q$ such that $P \oplus Q$ is a free $R$-module.
Now if the sequence splits, then $B \cong A \oplus P$, Does that imply that $B$ is a free module? I do not think so, that $B$ is a free module? if not then how do I show the existence of a module $R-$module $Q$ such that $P \oplus Q$
No, if $0\to A\to B\to P\to 0$ is exact and $P$ is projective then $B$ is not necessarily free. But we know that all short exact sequences with $P$ at the end split, so we can just choose one where $B$ is free. So let $(x_i)_{I}$ be any generating system of $P$ and let $K$ be the kernel of $R^{(I)}\to P$, $e_i\mapsto x_i$. Then we have the short exact sequence $$0\to K\to R^{(I)}\to P\to0$$ which splits by assumption, i.e. $R^{(I)}\cong K\oplus P$.
(Here $R^{(I)}=\bigoplus_{i\in I}R$)