To solve for \(\theta\) and \(P\), we resolve the forces horizontally and vertically, using the fact that the forces are in equilibrium.

Resolving horizontally:

\(3 \cos 60^{\circ} = 2 \cos \theta\)

\(\frac{3}{2} = 2 \cos \theta\)

\(\cos \theta = \frac{3}{4}\)

\(\theta = \cos^{-1}\left(\frac{3}{4}\right) \approx 41.4^{\circ}\)

Resolving vertically:

\(P = 3 \sin 60^{\circ} + 2 \sin \theta\)

\(P = 3 \times \frac{\sqrt{3}}{2} + 2 \times \sin 41.4^{\circ}\)

\(P \approx 3.92 \text{ N}\)