∠BAD = 75 ....angle on the center is twice the angle on the circumference
∠BPD = 180-75 = 105 .......because BADP is a cyclic quadrilateral 

∠BPD =∠ BCD ....because these angles are on the same segment

i hope it helps........ 
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As DOB is twice the angle of DAB
taking ABCD as cyclic quadrilateral
    DAB+DCB= 180°     (∵ opp.angles of a cyclic quadrilateral is supplimentary)
DCB=DPB               (∵ angles formed on the same segment of the circle                                                                                are equal) 
∴ DPB=105
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