I can not share the figure with you since I am a new member in this forum and by its rule, I need to post at least 10 times before I can share. So, let me explain this verbally. Draw a figure. Join SP. Assume two triangles - triangle QRS and PQS.

In triangle QRS, angle R = 90, and QR = RS = 5. So $area of triangle QRS = (1/2)*5*5$. Also, notice angle SQR = angle RSQ = 45, since it is a right isosceles triangle. Moreover, QS = 5\sqrt{2} by applying Pythagoras ratios in such a triangle.

In triangle PQS, angle PQS = 90, since 45° angle SQR is excluded from 135° angle PQR. Now PQ = 5, and QS = 5\sqrt{2}. So $area of triangle PQS = (1/2)* 5 * 5\sqrt{2}$

Add the areas of both triangle QRS and PQS.
C is the correct answer.