The correct option is B 62 cm
Let us assume BQ=x cm
Now, AQ=AB−BQ=(15−x) cm
By Theorem- Tangents drawn from an external point to the circle are equal in lengths.
∴DS=DP=4 cm,
CS=CR=12 cm,
BR=BQ=x cm, and
AP=AQ=(15−x) cm
Now, perimeter of a quadrilateral ABCD=AB+BC+CD+AD=AB+(BR+CR)+(CS+DS)+(AP+DP)=15+(x+12)+(12+4)+((15−x)+4)=15+12+12+4+15+4=62 cm
∴ Perimeter of a quadrilateral ABCD=62 cm