Option B: 50°
Since OA ⊥ PA and OB ⊥ PB (radius ⊥ tangent), in quadrilateral OAPB:
∠APB = 360° − 90° − 90° − 130° = 50°
In quadrilateral OAPB, the four angles sum to 360°. Angles OAP and OBP are each 90° (Theorem 10.1). So ∠APB = 360° − 130° − 90° − 90° = 50°. This is a standard application of the tangent-radius perpendicularity theorem. Note that ∠APB + ∠AOB = 180° (they are supplementary), which is a quick shortcut: 180° − 130° = 50°.