A metal M reacts with nitrogen gas to afford $$M_3N$$, $$M_3N$$ on heating at high temperature gives back M and on reaction with water produces a gas B. Gas B reacts with an aqueous solution of CuSO$$_4$$ to form a deep blue compound. M and B respectively are:

First we examine the information that the unknown metal $$M$$ reacts with nitrogen gas to give a nitride whose formula is written as $$M_3N$$.

Whenever we write an ionic nitride in the form $$M_3N$$ we are implying that the three atoms of the metal together supply the same total positive charge that the one nitride ion $$N^{3-}$$ carries as negative charge. Therefore each atom of the metal must provide a charge of $$+1$$ because

$$\text{Total positive charge} = 3 \times (+1) = +3$$
$$\text{Charge on }N^{3-} = -3$$

and the compound is electrically neutral. So the valency of the metal has to be $$+1$$. The only common main-group metal that fulfills two conditions - being monovalent and forming an isolable solid nitride of the stoichiometry $$M_3N$$ - is lithium. Sodium and the heavier alkali metals do not form a stable solid nitride under ordinary conditions, while group-2 metals would have formula $$M_3N_2$$ and not $$M_3N$$. Hence we are already strongly pointed toward

$$M = Li$$.

Let us write the preparative reaction explicitly:

$$6Li \;+\; N_2 \;\rightarrow\; 2Li_3N.$$

Now the question states that $$M_3N$$ on heating at high temperature “gives back” the metal $$M$$. For lithium nitride this decomposition can be represented as

$$2Li_3N \;\xrightarrow[\text{high }T]{}\; 6Li \;+\; N_2.$$

This step is perfectly consistent with the description.

The nitride is next allowed to react with water. The general hydrolysis of an ionic nitride is

$$M_3N \;+\; 3H_2O \;\rightarrow\; 3MOH \;+\; NH_3.$$

Substituting $$M = Li$$ we obtain

$$Li_3N \;+\; 3H_2O \;\rightarrow\; 3LiOH \;+\; NH_3.$$

The gas produced, designated in the problem statement as $$B$$, is therefore

$$B = NH_3 \;(\text{ammonia}).$$

The final piece of experimental evidence says that gas $$B$$ reacts with an aqueous solution of $$CuSO_4$$ to give a deep-blue compound. Ammonia is well known to form the tetraammine copper(II) complex, which has an intense deep-blue colour. The reaction may be written as

$$CuSO_4 \;+\; 4NH_3 \;\rightarrow\; [Cu(NH_3)_4]SO_4,$$

and the complex $$[Cu(NH_3)_4]^{2+}$$ in solution is responsible for the characteristic colour.

Because every piece of the description matches lithium and ammonia, the pair $$(M,\,B)$$ must be

$$M = Li,\qquad B = NH_3.$$

Looking at the options supplied, this corresponds to Option A.

Hence, the correct answer is Option A.