This is Exercise 2.2.2. from Weibel's An Introduction to Homological Algebra.
Suppose $A$ is an abelian category, if $A$ has enough projectives, then so does the category $Ch(A)$ of chain complex over $A$.
Weibel's hint is to use the fact: a chain complex $P$ ($P$ means each $P_n$ is a projective object in $A$) iff it's a split exact complex of projectives. And I have proven this. (For reference: one can find the two directions of the proof here and here.)
Let $B$ be a chain complex over $A$. Then for each $B_n$ we have a $P_n$ such that $P_n\rightarrow B_n$ is surjective.
I don't know how to get a differential $d_n: P_n\rightarrow P_{n-1}$ s.t. complex $P$ is split exact. I tried use the definition of $P_n$ to get a map from $P_n$ to $P_{n-1}$. But I don't know what to do next.
Best Answer
Let $M_{\bullet}$ be some chain complex. Choose epic maps $f_n:P_n\to M_n$ from projective objects $P_n$. Define $P_{\bullet}$ with the $n$th object $P_n\oplus P_{n+1}$ and define $d^P(a,b)=(0,a).$ Then $P_{\bullet}$ is split exact. Define an epic chain map $f:P_{\bullet}\to M_{\bullet}$ by $f(a,b)=f_na+d^Mf_{n+1}b$. It's a chain map because $$ d^Mf(a,b)=d^M(f_na+d^Mf_{n+1}b)=d^Mf_na=f(0,a)=fd^P(a,b). $$