Let PT and PQ are two tangents from external point P to a circle with center O

In △OPT and △OQT

OP = OQ

[radii of same circle]

OT = OT

[common]

PT = PQ

[Tangents drawn from an external point are equal]

△OPT ≅ △OQT

[By Side - Side - Side Criterion]

∠POT = ∠QOT

[Corresponding parts of congruent triangles are congruent]

i.e. Assertion is true

Now,

Consider a circle circumscribed by a parallelogram ABCD, Let side AB, BC, CD and AD touch circles at P, Q, R and S respectively.

As ABCD is a parallelogram

AB = CD and BC = AD

[opposite sides of a parallelogram are equal] [1]

Now, As tangents drawn from an external point are equal.

We have

AP = AS

[tangents from point A]

BP = BQ

[tangents from point B]

CR = CQ

[tangents from point C]

DR = DS

[tangents from point D]

Add the above equations

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

AB + CD = AS + DS + BQ + CQ

AB + CD = AD + BC

AB + AB = BC + BC [From 1]

AB = BC …[2]

From [1] and [2]

AB = BC = CD = AD

And we know,

A parallelogram with all sides equal is a rhombus

So, reason is also true, but not a correct reason for assertion.

Hence, B is correct option .