1. Write the given half-reactions

We are given:

$\text{(i)}\quad Ag + I^- \rightarrow AgI + e^- \qquad E^\circ = 0.152\,V$

$\text{(ii)}\quad Ag \rightarrow Ag^+ + e^- \qquad E^\circ = -0.800\,V$

We need to find $\log K_{sp}$ for:

$AgI(s) \rightleftharpoons Ag^+ + I^-$

2. Obtain the reaction for formation of $AgI$ from $Ag^+$ and $I^-$

Reverse reaction (ii):

$Ag^+ + e^- \rightarrow Ag \qquad E^\circ = +0.800\,V$

Now add it to reaction (i):

$Ag + I^- \rightarrow AgI + e^-$ $Ag^+ + e^- \rightarrow Ag$

On adding, $Ag$ and $e^-$ cancel:

$Ag^+ + I^- \rightarrow AgI$

For this reaction,

$E^\circ = 0.152 + 0.800 = 0.952\,V$

3. Relate $E^\circ$ to equilibrium constant

For the reaction

$Ag^+ + I^- \rightarrow AgI$

number of electrons transferred is $n=1$.

Using

$E^\circ = \frac{0.059}{n} \log K$

we get

$0.952 = 0.059 \log K$

So,

$\log K = \frac{0.952}{0.059} \approx 16.13$

Here $K$ is the equilibrium constant for:

$Ag^+ + I^- \rightarrow AgI$

But

$K = \frac{1}{K_{sp}}$

because

$AgI \rightleftharpoons Ag^+ + I^-$

has equilibrium constant $K_{sp}$.

Therefore,

$\log K_{sp} = -\log K = -16.13$

4. Final answer

$\boxed{\log K_{sp} = -16.13}$

So the correct option is D.