Consider a tuning fork as a source of sound which is moving towards the stationary pipe with velocity $u$.

According to the Doppler effect ,

$f\text{'}=f_s\left(\frac{v-v_o}{v-v_s}\right)$

Where,

$v=$is speed of sound

$v_s=$is velocity of source

$v_o=$is velocity of observer

$f^{\text{'}}=$is appeared frequency

$f_s=$is actual frequency

So the appeared frequency of the sound will be,

 $f\text{'}=f_s\left(\frac{v-0}{v-u}\right)$

Since pipe is closed at one end so it will be like a close organ pipe and we know that close organ pipe has only odd harmonics.

And for resonance, appeared frequency should match with any of the natural frequency of the closed organ pipe.If fundamental frequency of the closed organ pipe is $f_0$, then its natural frequencies will be $f_0,3f_0,5f_0....\left(2n-1\right)f_0$ 

So for resonance,

$f_s\left(\frac{v}{v-u}\right)=\left(2n-1\right)f_0$

$f_s=\left(1-\frac{u}{v}\right)\left(2n-1\right)f_0$

Now from options$\left(A,B,C\right)$

$\because u=0.8\mathrm{v}$

So,

$f_s=\left(1-0.8\right)\left(2n-1\right)f_0$

$f_s=\frac{2n-1}{5}{f}_0$

$n=1, f_s=\frac{f_0}{5}$

$n=2, f_s=\frac{3f_0}{5}$

$n=3, f_s=\frac{5f_0}{5}=f_0$

$n=4, f_s=\frac{7f_0}{5}$

$n=5, f_s=\frac{9f_0}{5}$

$n=6, f_s=\frac{11f_0}{5}$

So option (A) is correct, (B) and (C) are incorrect

Now from option$\left(D\right)$

$\because u=0.5\mathrm{v}$

So,

$f_s=\left(1-0.5\right)\left(2n-1\right)f_0$

$f_s=\left(\frac{2n-1}{2}\right)f_0$

$n=1, f_s=\frac{f_0}{2}$

$n=2, f_s=\frac{3f_0}{2}=1.5f_0$

$n=3, f_s=\frac{5f_0}{2}$

Hence option $\left(D\right)$ is correct.