Given that,

AB ‖ CD

Let, EF be the parallel line to AB and CD which passes through P.

It can be seen from the figure that alternate angles are equal

∠ABP = ∠BPF

∠CDP = ∠DPF

∠ABP + ∠CDP = ∠BPF + ∠DPF

∠ABP + ∠CDP = ∠DPB

Hence, proved