Property: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

By the above property,

AP = AS (tangent from A)

BP = BQ (tangent from B)

CR = CQ (tangent from C)

DR = DS (tangent from D)

Now we add above 4 equations,

AP + BP + CR + DR = AS + BQ + CQ + DS

⇒ AB + CD = AD + BC

[∵ AP + BP = AB

CR + DR = CD

AS + DS = AD

BQ + CQ = BC]

Hence, the right option is AB + CD = AD + BC