Let Q be the point in which the circumcircle of BCD meets again the line AB. Then ∠QEA=∠QBC=∠ECP, hence EQ∥PC. Moreover, ∠ECP=∠ECD implies QD is parallel to EC, hence EQ=CD=CP. It follows that EQCP is a parallelogram, which leads to the conclusion.

Second solution. (Given in the contest.) If {O}=BD∩AC, it is easy to prove that BA and BO are isogonal in the angle ∠EBC, therefore, in order to show that BA is the median, one has to prove that BO is the symmedian. This follows readily by computation, using Steiner's Theorem.
Indeed, ∠EBA=∠BAC−∠BEA=180∘−∠ABC−∠ACB−∠BDC=180∘−∠ECD−∠ECB−∠BDC=∠CBD.
Triangles OEB and ODC are similar, hence ODEO=CDEB. (1)
Triangles OCB and ODE are also similar, therefore ODOC=EDBC. (2)
We have ∠ADC=∠APC=∠PBC+∠PCB=∠ACB=∠EDB, which shows that the triangles ACD and EBD are similar. Hence ED=ACEB⋅AD and, as AD=BCCD⋅AC (from ΔACB∼ΔAPC∼ΔADC), it follows that ED=BCEB⋅CD.
From (1) and (2) we obtain OCEO=CD⋅BCEB⋅ED=(BCEB)2, which shows that BO is indeed the symmedian.
Third solution. (Given in the contest by Andrei Mărginean.) Let a=∠EBA, x=∠ABC. We have that ∠DCE=∠ACP=x and ∠EBC=a+x. As B, C, D, E are concyclic, it follows that ∠CDE=180∘−a−x, ∠DEC=180∘−∠EDC−∠DCE=a=∠EBA. But AE is the perpendicular bisector of [PD], hence ∠PEA=∠DEC=∠EBA, which shows that the triangles APE and AEB are similar. It follows that AE2=AB⋅AP. But triangles ACP and ABC are also similar, hence AC2=AP⋅AB=AE2, and the conclusion follows immediately.