Widzimy, że $PS\parallel BC$, więc $\left|\sphericalangle APS\right|=\left|\sphericalangle ABC\right|$ oraz $\left|\sphericalangle PSA\right|=\left|\sphericalangle BCA\right|$.

$△APS\sim △ABC$

Więc $\left|AS\right|=\left|PS\right|=2$

$\left|\sphericalangle CSR\right|=45^{\circ }$

$\left|CR\right|=2$ 

$\left|CS\right|=2\sqrt{2}$

$\left|AC\right|=\left|BC\right|=2+2\sqrt{2}$

$\left|BQ\right|=\left|BC\right|-\left|CR\right|-\left|QR\right|$

$\left|BQ\right|=2+2\sqrt{2}-4=2\sqrt{2}-2$

$P_{△PBQ}=\frac{2\cdot \left(2\sqrt{2}-2\right)}{2}=2\sqrt{2}-2$