For this question, I applied trig Ceva to get $\frac{\sin{80^\circ}}{\sin{20^\circ}} \cdot \frac{\sin{10^\circ}}{\sin{30^\circ}} \cdot \frac{\sin{\angle CBP}}{\sin{\angle ABP}} = 1$. However, I am not sure how to simplify all of the sines. I have tried double angle, but I'm not sure if that helps. Could I have a hint? Thanks.