Multiply the right side of your equation with $(\lambda I-P)$. Using $P^2=P$ you can verify that the product is $I$ for all $\lambda\ne 0,1$. It follows that $\lambda I-P$ is nonsingular for all $\lambda\ne 0,1$; therefore $0$ and $1$ are the only possible eigenvalues.
I don't know whether this equation has a name. You can "reinvent" it by looking at a two-dimensional situation where $P: (x,y)\mapsto (x,0)$ is the projection onto the $x$-axis. In this case $I-P$ is the projection onto the $y$-axis. You then have to answer the following question: How can I get back the point $z:=(x,y)$ when I'm given the point $w:=(\lambda I-P)z=(\lambda x-x,\lambda y)$?