ABCD is the rhombus with $\mathrm{B}\hat{\mathrm{A}}\mathrm{D}=60°$                            

The diagonals cut at O at right angle.

Take OC as the axis of x and OD as the axis of y.

Given OD = OB = 1.

$\therefore$ we have OA = OD cot 30° = $\sqrt{3}$  = OC

The radius OM = OA sin 30° = $\frac{\sqrt{3}}{2}$

$\therefore$ the circle has equation x² + y² = $\frac{3}{4}$

A, B, C, D are the points $(-\sqrt{3},0),(0,-1),(\sqrt{3},0),(0,1)$

If P be the point (α, β), then

|PA|² + |PB|² + |PC|² + |PD|² is equal to

${(\mathrm{\alpha }+\sqrt{3})}^2+{\mathrm{\beta }}^2+{\mathrm{\alpha }}^2+{(\mathrm{\beta }+1)}^2+{(\mathrm{\alpha }-\sqrt{3})}^2+{\mathrm{\beta }}^2+{\mathrm{\alpha }}^2+{(\mathrm{\beta }-1)}^2$

= 2 (α² + 3) + 2α² + 2(β² + 1) + 2β²

= 4(α² + β²) + 8

= 3 + 8 = 11 since α² + β² =$\frac{3}{4}$ or 4(α² + β²) = 3.