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Explanation:

Let the two circles with centres P and Q intersect at M and N.

Quadrilateral PMQN is a square.

m ∠MPN = m ∠MQN = 90°

The area common to both the circles = 2(Area of sector P-MN – Area of ∆PMN) 

$=2\left[\left(\frac{90}{360}\times \pi \times 1^2\right)-\left(\frac{1}{2}\times 1^2\right)\right]$

$=\frac{\pi }{2}-1$

Hence, option (b).