Make ∠ pad = 10 in ∠PAC and connect PC to D and BD.

Then ∠ DAC = ∠ PAC-∠ Dad = 40- 10 = 30.

∫∠PCA = 30

∴∠DAC=∠PCA

∴DA=DC

At △ABD and △CBD,

BA=BC

DA=DC

BD=BD

∴△ABD≌△CBD(SSS)

∴∠ABD=∠CBD=40

∠∠PDA =∠DAC+∠PCA = 60

∠PDB=∠DCB+∠CBD=60。

∴PD split equally ∠ ADB;

And ∠ pad = ∠ PAB = 10, that is, PA divided by ∠BAD, then point P is ∠ ⊿ADB's center.

∴∠ABP=∠DBP=20，∠PBC=60，

∠BPC = 180-∠PCB-∠PBC = 100。

I hope I can help you!