Show that if $0\rightarrow N\xrightarrow{\tau} P\xrightarrow{\epsilon} A\rightarrow 0$ and $0\rightarrow M\xrightarrow{\tau'} Q\xrightarrow{\epsilon'} A\rightarrow 0$ are exact sequences with $P,Q$ being projective modules then $P\oplus M\cong Q\oplus N$.
Consider surjective map $\epsilon$, as $Q$ is projective module and we have map $\epsilon'$ there exists $\eta:P\rightarrow Q$ such that $\epsilon'\circ \eta=\epsilon$
Moreover, $\epsilon'\circ (\eta\circ\tau)=(\epsilon'\circ \eta)\circ\tau=\epsilon\circ\tau=0$. Thus, $\eta\circ\tau$ is in the kernel of $\epsilon'$ and thus, has to be in the image of $\tau'$. So, we have a map $\eta':N\rightarrow M$ such that $\tau'\circ\eta'=\eta\circ\tau$
We then have a following diagram
$$\begin{array}{c} 0 & \rightarrow & N & \xrightarrow{\tau} & P & \xrightarrow{\epsilon} & A & \rightarrow & 0 \\ \downarrow{} & & \downarrow{\eta'} & & \downarrow{\eta} & & \downarrow{id} \\ 0 & \rightarrow & M & \xrightarrow{\tau'} & Q & \xrightarrow{\epsilon'} & A & \rightarrow & 0\\ \end{array}$$
From an exercise in Hilton Stammbach, Chapter I, exercise $3.4$ as map in last column is an isomorphism, we have a short exact sequence
$$0\rightarrow N\rightarrow P\oplus M\rightarrow Q\rightarrow 0$$
Now, as $Q$ is projective module, any surjective map onto $Q$ splits, so above short exat sequence is a split exact sequence, we then have $P\oplus M=Q\oplus N$.
Let me know if this is correct.