A source, approaching with speed 𝑢 towards the open end of a stationary pipe of length 𝐿, is emitting a sound of frequency $$f_s$$. The farther end of the pipe is closed. The speed of sound in air is v and $$f_0$$ is the fundamental frequency of the pipe. For which of the following combination(s) of u and $$f_s$$, will the sound reaching the pipe lead to a resonance?

For a pipe that is closed at one end and open at the other, only the odd harmonics are allowed.
The resonant frequencies are

$$f_n = (2n-1)\,f_0,\qquad n = 1,2,3,\ldots$$

The source is moving toward the open end of the pipe with speed $$u$$. For a moving source and a stationary observer (the pipe), the apparent frequency is given by the Doppler formula

$$f' = \frac{v}{\,v-u\,}\,f_s \tag{1}$$

All the options have $$u = 0.8\,v$$, so from (1)

$$f' = \frac{v}{v-0.8v}\,f_s = \frac{v}{0.2v}\,f_s = 5\,f_s \tag{2}$$

For resonance, the frequency that actually enters the pipe, $$f'$$, must equal one of the allowed frequencies $$f_n$$:

$$5\,f_s = (2n-1)\,f_0 \tag{3}$$

Divide by $$f_0$$ to get a condition on the ratio $$f_s/f_0$$:

$$\frac{f_s}{f_0} = \frac{2n-1}{5} \tag{4}$$

Now check each option.

$$f_s = f_0 \;\Longrightarrow\; \frac{f_s}{f_0}=1$$
From (4): $$1 = \frac{2n-1}{5}\;\Longrightarrow\;2n-1 = 5\;\Longrightarrow\;n = 3$$ (an integer). Hence resonance occurs.

$$f_s = 2f_0 \;\Longrightarrow\; \frac{f_s}{f_0}=2$$
Then $$2 = \frac{2n-1}{5}\;\Longrightarrow\;2n-1 = 10\;\Longrightarrow\;n = 5.5$$ (not an integer). No resonance.

$$f_s = 0.5f_0 \;\Longrightarrow\; \frac{f_s}{f_0}=0.5$$
Then $$0.5 = \frac{2n-1}{5}\;\Longrightarrow\;2n-1 = 2.5\;\Longrightarrow\;n = 1.75$$ (not an integer). No resonance.

$$f_s = 1.5f_0 \;\Longrightarrow\; \frac{f_s}{f_0}=1.5$$
Then $$1.5 = \frac{2n-1}{5}\;\Longrightarrow\;2n-1 = 7.5\;\Longrightarrow\;n = 4.25$$ (not an integer). No resonance.

Only Case A satisfies the integral value requirement for $$n$$, so resonance is obtained only for that choice.

Option A which is: u = 0.8v and $$f_s = f_0$$