Indija TS 2012

Extend $AD$ to $E$ such that $PE=PB$. Join $EB$ and $EC$.



$$\angle BPE=\angle BAC\text{ and }\frac{PB}{PE}=\frac{AB}{AC}=1.$$

Hence it follows that $CAB$ is similar to $EFB$. Thus $\angle PEB$. This shows that $A$, $B$, $E$, $C$ are concyclic. In turn we obtain $$\angle AEC=\angle ABC=\angle ACB=\angle AEB.$$

We conclude that $AE$ bisects $\angle BEC$.

Let $M$ be the mid-point of $BE$. Join $PM$ and $PC$. Since $EA$ bisects $\angle BEC$, we have $$\frac{CE}{EB}=\frac{CD}{DB}=\frac{CD}{2CD}=\frac{1}{2}.$$

Thus $CE=EB/2=EM$. We also observe that $\angle MEP=\angle CEP$. This shows that $CEP$ is congruent to $MEP$. This implies that $$\angle BAC=\angle BPE=2\angle MPE=2\angle DPC.$$