Serbian Mathematical Olympiad

Solution:

We carry out the proof in the case when $\angle BAC \leq 90^\circ$. The other case is analogous.
Let the lines $DE$ and $BC$ meet at $L$. The quadrilateral $CDPL$ is cyclic because $\angle PDL = \angle PCL$, from which we get $\angle PLE = \angle PCD = \angle BCA = \angle DAC = \angle DBE = \angle PBE$, so the quadrilateral $BPEL$ is also cyclic. From these two cyclicities we finally obtain $\angle PEB = \angle PLB = \angle PDC = \angle DBA = \angle DEA$.

Solution:

Second solution. Let $P'$ be a point on $BD$ such that $\angle DEA = \angle PEB$.
By the law of sines, $\frac{BP}{DP} = \frac{BP}{CP} \cdot \frac{CP}{DP} = \frac{\sin \angle BCP}{\sin \angle CBD} \cdot \frac{\sin \angle CDB}{\sin \angle PCD}$. Analogously, $\frac{BP'}{DP'} = \frac{\sin \angle BEP'}{\sin \angle EBD} \cdot \frac{\sin \angle EDB}{\sin \angle P'ED}$. Since $\angle BCP = \angle EDB$, $\angle CBD = \angle P'ED$, $\angle CDB = \angle BEP'$, and $\angle PCD = \angle EBD$, it follows that $\frac{BP}{DP} = \frac{BP'}{DP'}$, hence $P \equiv P'$.