Resolve forces perpendicular to the plane:

\(R + P \sin 30^{\circ} = 0.12g \cos 40^{\circ}\)

Using \(F = \mu R\), where \(\mu = 0.32\):

\(F = 0.32R\)

For the particle about to slip down the plane:

\(P_{\text{min}} \cos 30^{\circ} + F = 0.12g \sin 40^{\circ}\)

For the particle about to slip up the plane:

\(P_{\text{max}} \cos 30^{\circ} - F = 0.12g \sin 40^{\circ}\)

Substitute for \(F\) and solve for \(P\):

\(P \cos 30^{\circ} = 0.12g \sin 40^{\circ} \pm 0.32 (0.12g \cos 40^{\circ} - P \sin 30^{\circ})\)

Solving these equations gives:

\(P_{\text{max}} = 1.04\)

\(P_{\text{min}} = 0.676\)

Thus, the set of possible values for \(P\) is:

\(0.676 \leq P < 1.04\)