Solid $AgCl$ dissolves in aqueous ammonia mainly because

Aammonia lowers the $K_{s p}$ of $AgCl$ itself
Bammonia reacts with $Cl X^{-}$ to give $NH X_{4} Cl$ precipitate
Cammonia increases the $[Cl X^{-}]$ of the solution
D $[Ag (NH X_{3}) X_{2}] X^{+}$ forms, lowering $[Ag X^{+}]$ so that the ionic product falls below $K_{s p}$

Answer: D

Ag X^{+}

is removed as the stable complex

[Ag (NH X_{3}) X_{2}] X^{+}

, so

Q = [Ag X^{+}] [Cl X^{-}] < K_{s p}

and more

AgCl

dissolves.

K_{s p}

depends only on temperature, and

Cl X^{-}

is not consumed.