Solving the problem in the following way
[tex]\overrightarrow{PC}[/tex] bisects ∠BPD.
Then, ∠BPC=∠DPC-----(1)
AD is a line , on which ray PB and PC lies.
→∠APB+∠BPC+∠CPD=180°----By linear pair axiom
→124°+∠DPC+∠DPC=180°------[Using 1]
→124°+2∠DPC=180°
→∠DPC=180°-124°
→2∠DPC=56°
Dividing both sides by,2 we get
→∠DPC=28°
