If $P\in\Hom(E)$ is a projection onto $F$ with kernel $G$, then $P^*\in\Hom(E^*)$ is a projection onto $\{x^*\in E^*: x^*|G=0\}$ with kernel $\{x^*\in E^*: x^*|F=0\}$.
Since $(uv)^*=v^*u^*$ and $P^2=P$ we infer that $P^*P^*=(PP)^*=P^*$ and thus $P^*$ is a projection onto $\im P^*=\ker(1-P^*)$. $x^*\in E^*$ is in the kernel of $P^*$ iff for all $x\in E$:
$$
0=P^*(x^*)(x)=x^*(Px)
$$
i.e. if and only if $x^*|\im P=x^*|F=0$. $x^*\in E^*$ is in the image $\im P^*$ iff $x^*$ is in the kernel of $1-P^*$ and we just proved that this holds if and only if $x^*|\im(1-P)=x^*|G=0$.