A galvanic cell is set up as Ag(s) | AgCl(s), saturated AgCl(aq) || Ag^+(0.10 M) | Ag(s) at 298 K. Use Ksp(AgCl)=1.8×10^-10 and 2.303RT/F=0.0592 V.

The left compartment has only the sparingly soluble salt, so $[\mathrm{Ag^+}]=\sqrt{K_{sp}}=1.34\times10^{-5}\ M$. This is a concentration cell; the electrode in the more concentrated $\mathrm{Ag^+}$ solution ($0.10\ M$) has the higher potential and is the cathode, so (B) is correct. $E=0.0592\log(0.10/1.34\times10^{-5})=0.0592\times3.873=0.23\ \mathrm{V}$, so 0.52 V (obtained by wrongly using $K_{sp}$ itself for $[\mathrm{Ag^+}]$) is wrong and (A) is wrong. Added chloride lowers the free $[\mathrm{Ag^+}]$ on the left through the common-ion effect, which raises the ratio of concentrations, so $E$ rises and (C) is correct. Ammonia forms $\mathrm{[Ag(NH_3)_2]^+}$ and also lowers free $[\mathrm{Ag^+}]$, so the EMF increases, not decreases; (D) is wrong.