"A" obtained by Ostwald's method involving air oxidation of NH$$_3$$, upon further air oxidation produces "B". "B" on hydration forms an oxoacid of Nitrogen along with evolution of "A". The oxoacid also produces "A" and gives positive brown ring test

In the Ostwald process, ammonia is catalytically oxidized to nitric oxide according to the reaction

$$4NH_3(g)+5O_2(g)\xrightarrow{Pt/Rh}4NO(g)+6H_2O(l).$$

Hence,

$$A=NO.$$

Nitric oxide is then oxidized by atmospheric oxygen to form nitrogen dioxide:

$$2NO(g)+O_2(g)\rightarrow2NO_2(g).$$

Therefore,

$$B=NO_2.$$

Nitrogen dioxide dissolves in water to produce nitric acid and nitric oxide:

$$3NO_2(g)+H_2O(l)\rightarrow2HNO_3(aq)+NO(g).$$

The nitric acid formed gives a positive brown ring test due to the presence of nitrate ions.

Thus,

$$A=NO,\qquad B=NO_2.$$

Hence, the correct answer is

$$\boxed{\text{Option C}}.$$