From ${\angle}\mathrm{BPC} = {\angle}\mathrm{APD}$, we obtain ${\angle}\mathrm{BPD} = {\angle}\mathrm{APC}$. Additionally, since PB=PC and PD=PA, triangles PBD and PCA are congruent, leading to ${\angle}\mathrm{PBD} = {\angle}\mathrm{PCA}$. Observing that BC is parallel to PD, we have ${\angle}\mathrm{CBD} = {\angle}\mathrm{BDP}$.So from $\angle PBD+\angle BDO $$={\angle}\mathrm{PBC} - {\angle}\mathrm{CBD} + {\angle}\mathrm{BDP} + {\angle}\mathrm{PDO} = {\angle}\mathrm{PBC} + {\angle}\mathrm{PDO} = {\angle}\mathrm{CPD} + {\angle}\mathrm{DPO} $ $= {\angle}\mathrm{OPC} = {\angle}\mathrm{OCP}$. And since ${\angle}\mathrm{OCP} = {\angle}\mathrm{OCA} + {\angle}\mathrm{PCA}$, we find ${\angle}\mathrm{BDO} = {\angle}\mathrm{OCA}$. Given that BD=AC and OD=OC, triangle BDO is congruent to triangle ACO. Therefore, we have BO = AO. Q.E.D.