Let $A(0,0),D(1,0)$. The lines $y=\tan 80^{\circ} \cdot x$ and $y=-\tan 60^{\circ}(x-1)$ cut in the point $B(\frac{\sin 60^{\circ}\cos 80^{\circ}}{\sin 40^{\circ}},2\sin 60^{\circ}\cos 40^{\circ})$. Then $C(\frac{\sin 40^{\circ}+\sin 20^{\circ}}{2\sin 40^{\circ}},2\sin 60^{\circ}\cos 40^{\circ})$. The lines $y=-\tan 80^{\circ}(x-1)$ and $y=\tan 50^{\circ} \cdot x$ cut in the point $P(\frac{\tan 80^{\circ}}{\tan 50^{\circ}+\tan 80^{\circ}},\frac{\tan 80^{\circ}\tan 50^{\circ}}{\tan 50^{\circ}+\tan 80^{\circ}})$. The slope of the line $BP\ :\ m_{BP}=-\frac{\sqrt{3}}{3}=-\tan 30^{\circ}$, so $\angle PBC=0^{\circ}$.

From Langley triangle $m(\widehat{PBD})=30^\circ$, thus $m(\widehat{PBC})=30^\circ$. Best regards, sunken rock