How to calculate the solubility product for PbS from galvanic potential?
The galvanic cell must be written like this:

$$\ce{Pb | PbS saturated || Pb(NO3)2 \pu{1 M} | Pb}$$

in order to have a positive  $\mathrm{EMF} = \pu{0.413 V}$.

Your approach is right with a slight error.
This is a concentration cell based on $\ce{Pb^2+}$. Since $\ce{Pb(NO3)2}$ dissociates completely, while $\ce{PbS}$ is a salt with a low solubility, the left semicell will be the anode and the right one the cathode.
So, we have the following semireactions:

$$
\begin{align}
 \ce{Pb^2+ + 2 e^- &amp;-&gt; Pb} &amp;\quad &amp;\text{(cathode)} \\
 \ce{Pb &amp;-&gt; Pb^2+ + 2 e-} &amp;\quad &amp;\text{(anode)}
\end{align}
$$

And for the anode we also have

$$\ce{PbS &lt;=&gt; Pb^2+ + S^2-}$$

where $[\ce{Pb^2+}] = [\ce{S^2-}] = \sqrt{K_\mathrm{sp}}$.
So the semicell potentials are:

$$
\begin{align}
E_\text{cathode} &amp;= E^\circ\\
E_\text{anode}   &amp;= E^\circ + \frac{0.059}{2}\log_{10}[\ce{Pb^2+}]\\
\end{align}
$$

Thus:

$$\pu{0.413 V} = E_\text{cathode} - E_\text{anode} = -\frac{0.059}{2}\log_{10}[\ce{Pb^2+}] \implies [\ce{Pb^2+}] = 10^{-14}$$

And

$$K_\mathrm{sp} = [\ce{Pb^2+}]^2 = 10^{-28}$$