In isosceles triangle abc, ∠a = 80°. the bisectors of the base angles, ∠b and ∠c, intersect at point p. ∠bpc = 50° 100° 130°

First, find the measurement of ∠B and ∠C on the iscoceles triangle
the sum of angles in triangle is 180°
∠B + ∠C + ∠A = 180°
∠B + ∠C + 80° = 180°
∠B + ∠C = 100°
m∠B and m∠C are the leg angles in the iscoceles triangle, so they have the same measurement
2∠B = 100°
∠B = 50°
∠C = ∠B = 50°

Second, draw bisector to ∠B and ∠C. See image attached.
Then determine the measurement of ∠PBC and ∠PCB
∠PBC = 1/2 × 50°
∠PBC = 25°

∠PCB = 1/2 × 50°
∠PCB = 25°

Third, look at the new triangle formed, it's ΔBPC
Determine the measurement of ∠BPC using the sum of angles in triangle
∠BPC + ∠PCB + ∠PBC = 180°
∠BPC + 25° + 25° = 180°
∠BPC + 50° = 180°
∠BPC = 180° - 50°
∠BPC = 130°

3x+9+4x+x=8x+9that's the answer for your question.