The question is the one in the title. Let $(P,d)$ be a bounded complex of projective $R$-modules. My attempt is to consider, since projective is summand of free, for every integer $i$ a free module $F^i=P^i\oplus K^i$, which clearly comes with inclusion of and projection onto $P^i$.
So, there are morphisms $j_i:P^i\to F^i$ and $\pi_i:F^i\to P^i$, and one can give to $F$ the structure of chain complex declaring $\partial^i: F^i\to F^{i+1}$ to be $j_{i+1}\circ d^{i}\circ\pi_i$. This clearly makes the projections into a chain morphism. However it seems to me that it doesn't induces isomorphism on cohomology, since these aren't even isomorphic: $$\operatorname{Ker}(\partial^i)=\operatorname{Ker\begin{pmatrix}d^i & 0 \\ 0 & 0 \end{pmatrix}}=\operatorname{Ker}d^i\oplus K^i,$$while $\operatorname{Im}(\partial^{i-i})\cong\operatorname{Im}(d^{i-1})$.
Another approach that I was trying is to actually use boundedness and start a proof by induction for a sufficiently small index, where everything is $0$, but I don't know how to define the next free module in the inductive step...
Any suggestion? Thanks in advance!