Prove both inclusions. If v belongs to the column space of I−P, then v=(I−P)u for some u and Pv=(P−P2)u=0. Thus v∈N(P).
Conversely, if Pv=0, then v=(I−P)v, which explicitly writes vas an image of a linear map under I−P. Hence v∈C(I−P). The two inclusions prove N(P)=C(I−P). No orthogonality assumption was used.