To find the minimum force \( P \) required to prevent the block from sliding down the inclined plane, we need to analyze the forces acting on the block and apply the conditions for equilibrium.

Here are the forces acting on the block:

• Weight \( W \), acting vertically downwards.
• Normal reaction \( R \), perpendicular to the inclined plane.
• Frictional force \( F_f = \mu R \), acting up the plane, opposing motion.
• Applied force \( P \), parallel to the inclined plane up the plane.

Resolving the weight \( W \) into components:

• Component parallel to the plane: \( W \sin \alpha \)
• Component perpendicular to the plane: \( W \cos \alpha \)

For equilibrium along the plane:

P + \mu R = W \sin \alpha (1)

For equilibrium perpendicular to the plane:

R = W \cos \alpha (2)

Substitute equation (2) into equation (1):

P + \mu (W \cos \alpha) = W \sin \alpha

Solve for \( P \):

P = W \sin \alpha - \mu W \cos \alpha

Thus, the minimum value of \( P \) required to maintain equilibrium is:

Correct Answer: \( P = W (\sin \alpha - \mu \cos \alpha) \)

By reviewing the options, it's clear that this expression matches one of the given options. This derivation ensures that the block remains stationary by balancing all forces along and perpendicular to the plane.