How to calculate the solubility product for PbS from galvanic potential? Knowing that at 25 °C the following galvanic cell:
  $$\ce{Pb~|~Pb(NO_3)_2~1M~||~PbS~saturated~|~Pb}$$ shows an $\mathrm{EMF} =0.413~\mathrm{V}$, find the $K_\mathrm{sp}$ of $\ce{PbS}$.


My Approach
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 cathode and the right one the anode.
So we have the following semireactions:
\begin{align}
 \ce{Pb^2+ + 2e- &amp;-&gt; Pb}  &amp;&amp; \text{(cathode)} \\
 \ce{Pb  &amp;-&gt; Pb^2+ +2e- } &amp;&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:
$$0.413~\mathrm{V} = E_\text{cathode} - E_\text{anode} = \frac{0.059}{2} \log_{10}\ce{[Pb^2+]} \Rightarrow \ce{[Pb^2+]} = 10^{14} $$
And: $$K_\mathrm{sp} = \ce{[Pb^2+]}^2 = 10^{28}$$

I'm sure that I'm wrong for a sign but I don't understand where is the error.
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

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