(D) For the stationary observer: Frequency $= 400\; Hz$,Wavelength $= 0.85\; m$,Speed of sound $= 350\; m s^{-1}$.
For the running observer: Frequency $\approx 411.76\; Hz$,Wavelength $= 0.85\; m$,Speed of sound $= 340\; m s^{-1}$.
$1$. For the stationary observer:
Since the source and observer are stationary relative to each other,the frequency remains $400\; Hz$. The wind blows towards the observer,so the effective speed of sound is $v_e = v + w = 340 + 10 = 350\; m s^{-1}$. The wavelength is $\lambda = v_e / f = 350 / 400 = 0.875\; m$.
$2$. For the running observer:
Here,the air is still,so the speed of sound is $340\; m s^{-1}$. The wavelength remains $\lambda = v / f = 340 / 400 = 0.85\; m$. The observer moves towards the source at $10\; m s^{-1}$,so the frequency heard is $f' = f(v + v_o) / v = 400(340 + 10) / 340 = 411.76\; Hz$.
Conclusion: The two situations are not identical because the wavelength and the effective speed of sound differ in both cases.