Let ABCD is a parallelogram circumscribing with a circle.   

Now since ABCD is a parallelogram.

So AB = CD and AD = BC

Again AP = AS .........1  (since AP and AS are two tangents drawn from external point)

Similarly

PB =BQ .......2

CR = CQ ......3

DR  = DS......4

Adding all 4 equations, we get

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

=> (AP + PB) + (DR + CR) = (AS + DS) + (BQ + CQ)

=> AB + CD = AD + BC

=> AB +AB = AD + AD   (since AD = BC and AB = CD)

=> 2AB = 2AD

=> AB = AD

Since AB and AD are adjacent sides of parallelogram.

Hense parallelogram ABCD is a rhombus.