(i) The total weight of the boxes is given by:

\(R = 2 \times 400 \times 10 = 8000 \text{ N}\)

The maximum frictional force between the lower box and the ground is:

\(F = \mu R = 0.75 \times 8000 = 6000 \text{ N}\)

Therefore, the boxes will remain at rest if \(P \leq 6000\).

(ii) For no sliding between the boxes, the maximum frictional force between them is:

\(F_{\text{max}} = 0.4 \times 4000 = 1600 \text{ N}\)

Using Newton's second law for the upper box:

\(400a \leq 1600\)

Thus, \(a \leq 4\).

For the entire system, using Newton's second law:

\(P - 6000 = 800 \times a\)

Substituting \(a = 4\):

\(P - 6000 = 800 \times 4\)

\(P = 9200\)

Therefore, the maximum possible value of \(P\) is 9200.