B

Step 1: First, we find the frequency of the sound wave after it reflects off the wall. The frequency of sound remains 256 Hz as it reflects off the wall since the frequency of a wave is determined by the source.
Step 2: The Source frequency is 256 Hz. When a sound source approaches a stationary observer, the frequency perceived by the observer is given by:
$$ f' = f \times \frac{v + v_o}{v - v_s} $$
where:
- $f'$ is the frequency heard by observer
- $f$ is the actual frequency of the source (256 Hz)
- $v$ is the speed of sound in air (approximately 343 m/s)
- $v_o$ is the speed of observer (0 m/s since stationary)
- $v_s$ is the speed of the source moving towards observer.
Step 3: The frequency increases as the source moves towards the observer.
Step 4: The observer hears beats, which occur when two frequencies are close but not equal. Assuming the speed of the source is such that the frequency difference leads to audible beats with the reflected sound, the phenomenon validates that both frequency increases due to motion and reflections from surfaces lead to beat frequencies heard by the observer.
Therefore, it demonstrates the Doppler effect resulting in beat frequency. Hence the correct answer is option B.