(i) The normal reaction force \(R\) on box B is given by:

\(R = (200 + 250)g = 4500 \text{ N}\)

Using the limiting equilibrium condition, \(P = \mu R\), we have:

\(3150 = \mu \times 4500\)

Solving for \(\mu\), we get:

\(\mu = \frac{3150}{4500} = 0.7\)

(ii) For box A not to slide on B, the frictional force must equal the force required to accelerate A:

\(0.2 \times 200g = 200a\)

Solving for \(a\), we find:

\(a = 2 \text{ m s}^{-2}\)

Thus, \(a \leq 2 \text{ m s}^{-2}\).

(iii) Applying Newton's second law to the system of A and B:

\(P - F = 450a\)

Where \(F\) is the frictional force between A and B:

\(F = 0.2 \times 200g = 400 \text{ N}\)

Thus, the maximum \(P\) is:

\(P_{\text{max}} = 3150 + 450 \times 2 = 4050 \text{ N}\)