Equation of given circles are
(x−1)2+(y−1)2=4 and (x−3)2+(y−3)2=4.

Hence, C1(1,1) and r1=2; C2(3,3) and r2=2
⇒PC1=PC2=2
Now, by distance formula,
C1C2=(3−1)2+(3−1)2=22+22=8
⇒PC12+PC22=C1C22
⇒∠C1PC2=2π (by converse of pythagoras theorem in △PC1C2)
Hence, area of quadrilateral PC1QC2=2×area of △PC1C2=2×area of △QC1C2
=2×21×2×2=4