Option B: $4\sqrt{2}$ cm
Since OQ ⊥ PQ (radius ⊥ tangent) and ∠QPR = 90°, so ∠QPO = 45°. In right △OQP: $OP = \dfrac{OQ}{\sin 45°} = \dfrac{4}{1/\sqrt{2}} = 4\sqrt{2}$ cm.
Since ∠QPR = 90° and OP bisects ∠QPR (centre lies on angle bisector of the two tangents), ∠QPO = 45°. With OQ = 4 cm (radius) and ∠OQP = 90°, use sin 45° = OQ/OP to get OP = 4√2 cm. Key theorems used: radius ⊥ tangent, and OP bisects the angle between the two tangents.