The products A and B in the following reactions, respectively are

For halogenated alkanes the nature of the ambident nucleophile and the metal ion that accompanies it (Na+ or Ag+) decides whether substitution occurs through the carbon-end or the hetero-atom-end of the nucleophile.

Case 1 (formation of A):
Reagent used: $$AgNO_2$$ (silver nitrite).
• Nitrite ion is an ambident nucleophile: $$^{-}O\!{-}N\!O$$ has a negatively charged oxygen while the nitrogen also possesses a lone pair.
• In the presence of $$Ag^{+}$$ the reaction proceeds via a polar SN1-type mechanism that involves formation of a carbocation. Because $$Ag^{+}$$ strongly bonds with the leaving halide (AgCl ppt), the $$R^+$$ intermediate is generated; recombination occurs predominantly through the nitrogen end of the nitrite ion, giving a nitroalkane $$R{-}NO_2$$.
Hence for 1-chloropropane the product is $$CH_3{-}CH_2{-}CH_2{-}NO_2$$ (1-nitropropane).

Case 2 (formation of B):
Reagent used: $$AgCN$$ (silver cyanide).
• Cyanide ion is also ambident: $$^{-}C\! \equiv \!N$$ can attack through carbon (to give nitrile) or through nitrogen (to give isocyanide).
• With $$Ag^{+}$$, as before, a carbocation is produced. Attack now favours the lone pair on nitrogen, so the isomer formed is an isocyanide (isonitrile) $$R{-}NC$$ rather than the usual nitrile $$R{-}CN$$ obtained with alkali metal cyanides.
Therefore from 1-chloropropane the product is $$CH_3{-}CH_2{-}CH_2{-}NC$$ (propyl isocyanide).

Thus,
A = $$CH_3{-}CH_2{-}CH_2{-}NO_2$$
B = $$CH_3{-}CH_2{-}CH_2{-}NC$$

Option D which is: $$CH_3{-}CH_2{-}CH_2{-}NO_2,\;CH_3{-}CH_2{-}CH_2{-}NC$$ is the correct answer.