Option A: $r\sqrt{2}$
Since PT is a tangent at T, $\angle OTP = 90°$. In right triangle OTP, $\angle POT = 45°$, so $\cos 45° = \dfrac{OT}{OP} = \dfrac{r}{OP}$, giving $OP = \dfrac{r}{\cos 45°} = r\sqrt{2}$.
The key property used is: tangent ⊥ radius at point of contact (Theorem 10.1), so ∠OTP = 90°. Then apply trigonometry (or Pythagoras) in right triangle OTP. With ∠POT = 45° and the remaining angle ∠TPO = 45°, the triangle is an isosceles right triangle, so $OP = OT\sqrt{2} = r\sqrt{2}$. Watch out for option B ($\sqrt{2r}$) — it is a common distractor; the correct form is $r\sqrt{2}$, not $\sqrt{2r}$.