$A$ is a $k$-algebra and $b$ is an idempotent in $Z(A)$. Let $P$ be a projective $A$-module. Then $bP$ is a projective $Ab$-module.
Here is my proof: $P$ is a direct summand of a free $A$-module and hence a direct summand of a free $Ab$-module since $A = A(1-b)\oplus Ab$ and $A(1-b) \cong Ab$. So since $bP$ is a direct summand of $P$, it's projective as a $Ab$-module.
Any verification would be appreciated!