Solution:

Let the circle have center $O$ and radius $R,$ with $A = (R, 0),$ $B = (0, R),$ $C = (-R, 0),$ $D = (0, -R),$ and $P = (R\cos\theta, R\sin\theta).$ Then $PA^2 = 2R^2(1 - \cos\theta)$ and $PC^2 = 2R^2(1 + \cos\theta),$ so $PA \cdot PC = 2R^2|\sin\theta| = 56.$ In the same way $PB \cdot PD = 2R^2|\cos\theta| = 90.$

Squaring and adding, $4R^4 = 56^2 + 90^2 = 11236,$ so $2R^2 = \sqrt{11236} = 106.$ The square has diagonal $2R,$ hence area $\frac{(2R)^2}{2} = 2R^2 = 106.$