Exercise 2.2.1 in Charles A. Weibel's book An Introduction To Homological Algebra How to prove that a chain complex is a projective object in $ {Ch} $ (chain complexes of $R$-modules) iff it is a split exact complex of projectives? 
A chain complex of projectives means a chain complex $\mathbb P_{\bullet}$ in which each $P_n$ is projective. 
Hint: To see that P must be split exact, consider the surjection $\operatorname{cone}(\operatorname{id})\to P[-1]$. To see that split exact complexes are projective objects, consider the special case $0 \to P_1 \cong P_0 \to 0 $
 A: I will give a more detailed hint on how to show a projective object $P$ of the category of chain complexes must be a split exact complex of projectives.  There is an exact sequence
$$ 0 \to P \to \operatorname{cone}(\operatorname{id}) \to P[-1] \to 0. $$
$P$ is projective, so $P[-1]$ is too, so this short exact sequence splits.   Thus $\operatorname{cone}(\operatorname{id}) \cong P \oplus P[-1]$.  But the cone is always exact, so $P$ is exact too.
The surjection you mentioned is $(p_{n-1}, p_n)\mapsto -p_{n-1}$, so the splitting $\phi : P[-1] \to \operatorname{cone}(\operatorname{id})$ has the form $p_{m-1} \mapsto (-p_{m-1}, \theta_{m-1}(p_{m-1}))$ where $\theta: P[-1] \to P$.  Now use the fact that $\phi$ is a map of chain complexes, and your knowledge of the differential on the cone, to see that $\theta$ is a chain contraction of the identity and therefore that $P$ is split exact.  Remember that the differential on $P[-1]$ is minus that on $P$.
Finally you need to show each $P_n$ is projective.  Since $P$ is split exact, $P_n=B_n\oplus B'_n$ and the differential is zero on $B_n$ and an isomorphism from $B'_n$ to $B_{n-1}$.  It's enough to show each $B_r'$ is projective, so suppose there is a module map $f:B_r' \to M$, and consider a module surjection $N \to M$ and the resulting surjection of stalk complexes
$$ (\cdots 0 \to N \to 0 \cdots  ) \to (\cdots 0 \to M \to 0 \cdots) $$
with $M$ and $N$ in degree $r$.
There is a map of complexes from $P$ to the second of these which is $f\oplus 0$ in degree $r$ and zero everywhere else.  By projectivity of $P$, this factors through the surjection of stalk complexes above. You can use this to get that $B_r'$ is projective.
