$V$ is a finitte dimension vector space. If for some $P\in End(V)$ we have $P=P^2$ then $P$ is an orthogonal projection $\iff$ $P$ is self-adjoint.
I can show that since $P$ is idempotent then $V=ker(P)\oplus Im(P)$ and $P$ is the projection on its image parallel to its kernel. If $P$ is also self-adjoint I easily get that $ker(P)$ is orthogonal to $Im(P)$ and therefore $P$ is an orthogonal projection.
I'm not quite sure how to proceed in the other direction, thoughts?
Edit - Answer (Thanks to Verdruss)
As shown above $\forall v,v'\in V$ we have a unique decomposition $v=u+w,\ v'=u'+w'$ $\ \ u,u'\in U;\ \ w,w'\in W$ where $U$ and $W$ are orthogonal.
$\left\langle P(v)\mid v'\right\rangle =\left\langle P(u+w)\mid u'+w'\right\rangle =\left\langle u\mid u'+w'\right\rangle =\left\langle u\mid u'\right\rangle +\left\langle u\mid w'\right\rangle =\left\langle u\mid u'\right\rangle $
$\left\langle v\mid P(v')\right\rangle =\left\langle u+w\mid P(u'+w')\right\rangle =\left\langle u+w\mid u'\right\rangle =\left\langle u\mid u'\right\rangle +\left\langle w\mid u'\right\rangle =\left\langle u\mid u'\right\rangle $
And therefore we can conclude $P$ is self-adjoint