The standard reduction potentials for two reactions are given below : 
$\begin{array}{l}{\mathrm{AgCl}}_{\left(\mathrm{s}\right)}+{\mathrm{e}}^{-}⟶{\mathrm{Ag}}_{\left(\mathrm{s}\right)}+{\mathrm{Cl}}^{-};{\mathrm{E}}^0=0.22\mathrm{V}\\ {\mathrm{Ag}}_{\left(\mathrm{aq}\right)}^{+}+{\mathrm{e}}^{-}⟶{\mathrm{Ag}}_{\left(\mathrm{s}\right)};{\mathrm{E}}^0=0.80\mathrm{V}.\end{array}$
The solubility product of AgCl under standard conditions of temperature (298 K) is given by

${\mathrm{AgCI}}_{\left(\mathrm{s}\right)}+{\mathrm{e}}^{-} \to {\mathrm{Ag}}_{\left(\mathrm{s}\right)}+{{\mathrm{Cl}}^{-}}_{\left(\mathrm{aq}\right)} {\mathrm{E}}^{\mathrm{o}} = 0.22 \mathrm{V}$
${{\mathrm{Ag}}^{+}}_{{\left(\mathrm{aq}\right)}_{ }}+{\mathrm{e}}^{-}\to {\mathrm{Ag}}_{\left(\mathrm{s}\right)}, {\mathrm{E}}^{\mathrm{o}} = 0.80 \mathrm{V}$
${\mathrm{Ag}}_{\left(\mathrm{s}\right)} \to {\mathrm{Ag}}_{\left(\mathrm{aq}\right)}^{+}+{\mathrm{e}}^{-}, {\mathrm{E}}^{\mathrm{o}} = -0.80 \mathrm{V}$
Adding equation (1) and (2) we get
${\mathrm{AgCI}}_{\left(\mathrm{s}\right)}\to {\mathrm{Ag}}_{\left(\mathrm{aq}\right)}^{+}+{\mathrm{CI}}_{\left(\mathrm{aq}\right)}^{-} {\mathrm{E}}^{\mathrm{o}} = -0.58\mathrm{V}$
from Nernst equation,
${\mathrm{E}}_{\mathrm{cell}} = {\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}-\frac{\mathrm{RT}}{\mathrm{nF}}\mathrm{lnk}$
$0 = -0.58-\frac{8.314\times 298\times 2.303}{1\times 96500}\mathrm{logk}$
$\log \mathrm{k} = \frac{0.58}{0.0591} = -9.81$
$\mathrm{k} = 1.5\times {10}^{-10}$
the solubility product of AgCI is $1.5\times {10}^{-10}$.