Since the system is in equilibrium, the tension \(T\) in the string is equal to the weight of the particle \(P\). Therefore, \(T = 3g\), where \(g\) is the acceleration due to gravity. Assuming \(g = 10 \text{ m/s}^2\), we have:

\(T = 3 \times 10 = 30 \text{ N}\).

For the normal component of the contact force on \(B\), resolve the forces perpendicular to the plane. The normal force \(R\) is given by:

\(R = (4g - T) \cos \alpha\).

Substitute \(T = 30 \text{ N}\) and \(\cos \alpha = 0.8\):

\(R = (4 \times 10 - 30) \times 0.8 = 10 \times 0.8 = 8 \text{ N}\).