(c) 25o
Exterior angle of triangle is equal to sum of 2 opposite interior angles.
As BC is straight line
∠PBD + 120° = 180°
∠PBD = 180° – 120°
∠PBD = 60°
As ΔPBD is isosceles triangle and PB = PD
∴ ∠PBD = ∠PDB
⇒ ∠PDB = 60°
In ΔPSR
With exterior angle ∠PSQ equal to sum of opposite interior angles
∠PDB = ∠DPC + ∠PCD
60° = x + 35°
x = 60° – 35°
x = 25°