First, calculate the normal reaction force R:

\(R = 0.6g \cos 21^{\circ} \approx 5.60 \text{ N}\)

Next, calculate the frictional force F using \(F = \mu R\):

\(F = 0.3 \times 5.60 \approx 1.68 \text{ N}\)

For the particle to be in equilibrium, the sum of forces parallel to the plane must be zero. Consider the case where the particle is on the verge of slipping down:

\(P + F = 0.6g \sin 21^{\circ}\)

\(P + 1.68 = 2.15\)

\(P = 2.15 - 1.68 = 0.470 \text{ N}\)

Now consider the case where the particle is on the verge of slipping up:

\(P - F = 0.6g \sin 21^{\circ}\)

\(P - 1.68 = 2.15\)

\(P = 2.15 + 1.68 = 3.83 \text{ N}\)

Thus, the least possible value of P is 0.470 N, and the greatest possible value of P is 3.83 N.