Let $AP,CD$ intersect in $X$ and $AB,DP$ intersect in $Y$. Then by triangle similarities $$\frac{AB}{CX}=\frac{BP}{CP}=\frac{BY}{CD}$$Let $C'\in AD$ be such a point that $CC'\parallel AP$. Then by Tales' theorem $$\frac{DC'}{AC'}=\frac{CD}{CX}$$Let $B'\in AD$ be such a point that $BB'\parallel DP$. Then by Tales' theorem $$\frac{DB'}{AB'}=\frac{BY}{AB}$$Hence $$\frac{DC'}{AC'}=\frac{DB'}{AB'}$$. Since $P$ lies on segment $BC$ both points $B',C'$ lie on segment $AD$. Thus above equality proves $B'=C'$.QED