(i) Resolve the forces horizontally and vertically:

Horizontal component: \(R_x = 40 \times \frac{24}{25} - 30 \times \frac{7}{25} = 30\)

Vertical component: \(R_y = 50 - 40 \times \frac{7}{25} - 30 \times \frac{24}{25} = 10\)

Magnitude of the resultant force:

\(R = \sqrt{R_x^2 + R_y^2} = \sqrt{30^2 + 10^2} = 31.6 \text{ N}\)

Direction of the resultant force:

\(\theta = \arctan \left( \frac{R_y}{R_x} \right) = \arctan \left( \frac{10}{30} \right) = 18.4^\circ\)

(ii) To make the resultant act in the positive \(x\)-direction, set \(R_y = 0\):

\(50 - P \times \frac{7}{25} - 30 \times \frac{24}{25} = 0\)

Solving for \(P\):

\(P = 40\)