For all $g\in G$, we have $\Psi(g)\in \Gl(E)$, so $\Psi(g)^{-1}\in \Gl(E)$ and $\Psi^d(g) = \Psi(g)^{-1*}\in \Gl(E^*)$ (this is indeed an invertible operator - its inverse is $\Psi(g)^*$). So we just have to show that $\Psi^d$ is a group homomorphism: For $g,h\in G$, we have $$ \Psi^d(gh) = \Psi(gh)^{-1*} = (\Psi(g)\Psi(h))^{-1*} = (\Psi(h)^{-1}\Psi(g)^{-1})^* = \Psi(g)^{-1*}\Psi(h)^{-1*} = \Psi^d(g)\Psi^d(h). $$ Thus, $\Psi^d$ is a representation of $G$ in $E^*$.