Resolve the forces perpendicular to the plane:

\(R = 12g \cos 25^{\circ} + P \sin 25^{\circ}\)

Resolve the forces parallel to the plane:

\(P \cos 25^{\circ} = F + 12g \sin 25^{\circ}\)

Since the particle is on the point of moving, the frictional force \(F\) is at its maximum value:

\(F = \mu R = 0.8R\)

Substitute \(F = 0.8R\) into the parallel force equation:

\(P \cos 25^{\circ} = 0.8R + 12g \sin 25^{\circ}\)

Substitute \(R = 12g \cos 25^{\circ} + P \sin 25^{\circ}\) into the equation:

\(P \cos 25^{\circ} = 0.8(12g \cos 25^{\circ} + P \sin 25^{\circ}) + 12g \sin 25^{\circ}\)

Simplify and solve for \(P\):

\(P \cos 25^{\circ} = 9.6g \cos 25^{\circ} + 0.8P \sin 25^{\circ} + 12g \sin 25^{\circ}\)

\(P \cos 25^{\circ} - 0.8P \sin 25^{\circ} = 9.6g \cos 25^{\circ} + 12g \sin 25^{\circ}\)

\(P(\cos 25^{\circ} - 0.8 \sin 25^{\circ}) = 9.6g \cos 25^{\circ} + 12g \sin 25^{\circ}\)

\(P = \frac{9.6g \cos 25^{\circ} + 12g \sin 25^{\circ}}{\cos 25^{\circ} - 0.8 \sin 25^{\circ}}\)

Substitute \(g = 9.8\):

\(P = \frac{9.6 \times 9.8 \times \cos 25^{\circ} + 12 \times 9.8 \times \sin 25^{\circ}}{\cos 25^{\circ} - 0.8 \times \sin 25^{\circ}}\)

\(P = 242\)