To solve this problem, we need to find the ratio of distances from station A to point P and station B to point P, where the two trains meet.

First, let's find the speeds of both trains. The train from station A starts at 9 AM and reaches station B at 2 PM, taking a total of 5 hours for the journey. Similarly, the second train starts at 11 AM and reaches station A at 3 PM, also taking 4 hours for the journey. This indicates that the distance between the two stations is covered by the trains in their respective times.

Let the distance from station A to station B be \(D\).

The speed of the first train \(= \frac{D}{5}\)

The speed of the second train \(= \frac{D}{4}\)

Next, we find the time taken by both trains from their starting points to meet at point P. Let this time be \(t\) hours for the first train and \((t - 2)\) hours for the second train (since it started 2 hours later at 11 AM).

Using the speed, the distances traveled by the two trains at the time they meet at point P:

Distance traveled by the first train \(= \frac{D}{5} \cdot t\)

Distance traveled by the second train \(= \frac{D}{4} \cdot (t - 2)\)

Since these distances add up to the total distance between stations A and B:

\(\frac{D}{5} \cdot t + \frac{D}{4} \cdot (t - 2) = D\)

To simplify, divide both sides by \(D\):

\(\frac{t}{5} + \frac{(t - 2)}{4} = 1\)

Clearing the fractions by multiplying the equation by 20:

\(4t + 5(t - 2) = 20\)

\(4t + 5t - 10 = 20\)

\(9t = 30\)

\(t = \frac{30}{9} = \frac{10}{3}\)

Thus, the first train takes \(\frac{10}{3}\) hours to reach point P. The distance it covers is:

\(\frac{D}{5} \times \frac{10}{3} = \frac{2D}{3}\)

Similarly, the second train takes:

\(\frac{10}{3} - 2 = \frac{10}{3} - \frac{6}{3} = \frac{4}{3}\) hours

The distance it covers in this time is:

\(\frac{D}{4} \times \frac{4}{3} = \frac{D}{3}\)

Thus, the ratio of the distances from station A to P and from station B to P is:

\(\frac{\frac{2D}{3}}{\frac{D}{3}} = \frac{2}{1}\)

Therefore, the correct answer is 2:1.