Let $O_{1}$ and $O_{2}$ be the centers of $\omega_{1}$ and $\omega_{2}$, respectively. Because $$\angle O_{2}AP+\angle O_{2}BP=90^{\circ}+90^{\circ}=180^{\circ}$$ quadrilateral $O_{2}APB$ is cyclic. But $O_{2}, A$, and $B$ lie on $\omega_{1}$, so $P$ lies on $\omega_{1}$ and $O_{2}P$ is a diameter of $\omega_{1}$. From the Pythagorean theorem on triangle $PAP_{2}$, we can calculate $AP=4\sqrt{6}$, so $\sin \angle AOP=\frac{2\sqrt{6}}{5}$ and $\cos \angle AOP=\frac{1}{5}$. Because $\triangle AOP$ and $\triangle BOP$ are congruent, we have $$\sin \angle APB=\sin 2 \angle AOP=2 \sin \angle AOP \cos \angle AOP=\frac{4\sqrt{6}}{25}$$ implying that $$[APB]=\frac{PA \cdot PB}{2} \sin \angle APB=\frac{192\sqrt{6}}{25}$$