Let PCD = x,
∫CP sharing ∠ACD,
∴∠ACP=∠PCD=x,PM=PN,
∫BP divided equally ∠ABC,
∴∠ABP=∠PBC,PF=PN,
∴PF=PM,
∫∠BPC = 40,
∴∠ABP=∠PBC=(x-40),
∴∠bac=∠acd-∠abc=2x-(x-40)-(x-40)= 80,
∴∠CAF= 100,
At Rt△PFA and Rt△PMA,
PA=PA,PM=PF,
∴Rt△PFA≌Rt△PMA,
∴∠FAP=∠PAC=50。
So the answer is: 50.