The external bisector of $$\angle B$$ and $$\angle C$$ of $$\triangle ABC$$ (where AB & AC extended to E and F respectively) meet at point P. If $$\angle BAC=100^\circ$$ , then the measure of $$\angle BPC$$

A

$$50^\circ$$

B

$$80^\circ$$

C

$$40^\circ$$

D

$$100^\circ$$