Show that $V=ker(P)+im(P)$ and $ker(P)\cap im(P)=\{0_v\}$
For all $x\in V$ we have
$$x=\underbrace{(x-p(x))}_{\in\ker p}+\underbrace{p(x)}_{\in im p}$$
and if $y\in \ker p\cap im p$ then $y=p(x)$ for some $x\in V$ so $$0=p(y)=p^2(x)=p(x)=y$$ hence
$$\ker p\cap im p=\{0\}$$
For $v \in V$ write $$v = (v-P(v)) + P(v) \in \ker P + Im P$$ since $P(v-P(v)) = P(v) - P(P(v)) = P(v) - P(v) = 0$