imc

When we see this problem, it practically screams similar triangles at us. Extend $OP$ to the left until it intersects lines $AD$ and $BC$ at point $E$. Triangles $EBO$ and $ECP$ are similar, and by symmetry, so are triangles $EAO$ and $EDP$. Then, the area of kite $EAOB$ is just $4\sqrt{2}\times 2$ and the area of kite $EDPC$ is $8\sqrt{2}\times 4$ (using our similarity ratios). The difference of these yields the area of hexagon $AOBCPD$ or $24\sqrt{2}⟹\boxed{24\sqrt{2}}$.