haloalkanes.

i think the reactions going to go via Benzyne mechanism bcoz. direct attack of NH2 at ortho position is highly unfavourable due to pumped up electron density at ortho by Methoxy group, after benzyne type intermediate forms sodamide+ammonia can add NH2 easily at meta

Okay let's do this question rigorously by mechanism. Firstly note that resonance effects are not taken in account when doing benzyne reactions. Only inductive effects are. This is because during the entire course of the reaction, the benzyne intermediate prevents any successful resonance in the ring.
So -OCH₃ has a +R effect which will not be counted. It has a -I effect.

Step I : The most acidic proton will be abstracted by sodamide. This will be the proton next to the carbon having Br due to combined -I effects, say. A minus charge is formed thus.

Step II : The minus charge knocks out -Br and forms the characteristic triple bond of benzyne. This is the slow step, the RDS of the reaction.

Step III : Attack of the nucleophile. This is a fast step, and it occurs equally on both sides of the triple bond, giving 50-50% products. There is no such major product formed in the reaction. The answer should be both (A) and (D).

Note : An acid base reaction could also occur at the carbon below -OCH₃, but
1. That proton would be less acidic as it is farther from Br(strongest inductor).
2. There is no scope of leaving groups. So the proton would just be given back(acid base reactions are fast and reversible).