因為,∠BPA = 180°-∠BAP-∠ABP = 70° = ∠BAP ,
所以,BA = BP ,可得：△ABP是等腰三角形；
過點A作AD⊥BC于D,交CP延長線于O,連接OB；
過點B作BE⊥CP于E,則點E在CO延長線上；
AD是等腰△ABC底邊上的高,可得：AD是BC的垂直平分線,
而且O在AD上,可得：OB = OC ,
∠OBC = ∠OCB = 30° ,
∠CBE = 90°-∠OCB = 60° ,
∠OBE = ∠CBE-∠OBC = 30° ；
因為,在△OBD和△OBE中,∠ODB = 90° = ∠OEB ,∠OBD = 30° = ∠OBE ,OB = OB ,
所以,△OBD ≌ △OBE ,
可得：OD = OE ,BD = BE ；
因為,在Rt△ABD和Rt△PBE中,AB = PB ,BD = BE ,
所以,△ABD ≌ △PBE ,
可得：AD = PE ；
因為,在△BOA和△BOP中,OA = AD-OD = PE-OE = OP ,BA = BP ,OB = OB ,
所以,△BOA ≌ △BOP ,
可得：∠OBA = ∠OBP = ½∠ABP = 20° ,
所以,∠PBC = ∠OBC-∠OBP = 10° .