Step-by-step Solution:

1. Identify the Source and the Observer:

   • The source of the sound is the whistle from the engine of train A.
   • The observers are the passengers inside train A.

2. Determine the Relative Motion:

   • Train A is moving at a speed of $v_A = 20 \mathrm{~m/s}$.
   • Since both the engine (source) and the passengers (observers) are on the same train (train A), they are moving with the same velocity.
   • Therefore, the relative velocity between the source and the observer is zero. $v_{rel} = v_{source} - v_{observer} = 20 \mathrm{~m/s} - 20 \mathrm{~m/s} = 0$.

3. Apply the Doppler Effect Principle:

   • The Doppler effect describes the change in frequency of a wave in relation to an observer who is moving relative to the wave source.
   • The general formula for the observed frequency ($f'$) is given by: $f' = f \left( \frac{v \pm v_o}{v \mp v_s} \right)$ where $f$ is the source frequency, $v$ is the speed of sound, $v_o$ is the observer's speed, and $v_s$ is the source's speed.
   • A Doppler shift (i.e., $f' \neq f$) occurs only if there is relative motion between the source and the observer.

4. Analyze the Outcome for Passengers in Train A:

   • Since the relative velocity between the engine of train A (source) and the passengers of train A (observers) is zero, there is no Doppler effect.
   • This means the frequencies observed by the passengers will be exactly the same as the frequencies emitted by the source.
   • The problem states that the source whistle emits frequencies ranging from $f_1 = 800 \mathrm{~Hz}$ to $f_2 = 1120 \mathrm{~Hz}$. The spread in frequency is $\Delta f = f_2 - f_1 = 320 \mathrm{~Hz}$.
   • Therefore, the passengers in train A will observe the same frequency range (800 Hz to 1120 Hz) and the same intensity distribution as emitted by the source.

5. Evaluate the Options:

   • The question asks for the best representation of the sound intensity distribution as observed by the passengers in train A.
   • We must look for the graph that shows the original, un-shifted frequency distribution, which spans from 800 Hz to 1120 Hz.
   • Option A: This graph shows the intensity distribution over the frequency range from 800 Hz to 1120 Hz. This matches the source distribution.
   • Option B and C: These graphs show the frequency range shifted to higher or lower frequencies, respectively. This would be the case if there were a Doppler effect, for instance, for an observer on train B or a stationary observer.
   • Option D: This graph shows a distribution with a significantly different frequency spread.

6. Conclusion:

   • The correct option is the one that represents the original sound distribution. The information about train B is extraneous and acts as a distractor.
   • Option A correctly depicts the frequency distribution from 800 Hz to 1120 Hz, which is what the passengers on train A would hear.

(Self-check: For passengers on train B, which is moving ahead of A, the source and observer are moving away from each other. The observed frequency would be lower, given by $f' = f \left( \frac{v - v_o}{v - v_s} \right) = f \left( \frac{340 - 30}{340 - 20} \right) = f \frac{310}{320}$. This would result in a frequency shift to lower values and a narrower spread, likely corresponding to option C. This confirms that the other options represent different physical scenarios.)