First, resolve the forces perpendicular to the plane to find the normal reaction \(R\):

\(R = 20g \cos 60^{\circ} = 100\).

The frictional force \(F\) is given by \(F = \mu R = 100\mu\).

Resolve the forces parallel to the plane for maximum and minimum values of \(P\):

For maximum \(P\), \(P_{\text{max}} = 20g \sin 60^{\circ} + F\).

For minimum \(P\), \(P_{\text{min}} = 20g \sin 60^{\circ} - F\).

Given \(P_{\text{max}} = 2P_{\text{min}}\), substitute the expressions:

\(20g \sin 60^{\circ} + F = 2(20g \sin 60^{\circ} - F)\).

Simplify and solve for \(\mu\):

\(20g \sin 60^{\circ} + 100\mu = 40g \sin 60^{\circ} - 200\mu\).

\(300\mu = 20g \sin 60^{\circ}\).

\(\mu = \frac{20g \sin 60^{\circ}}{300} = \frac{\sqrt{3}}{3} = 0.577\).