How does precipitate affect cell potential?
The solubility product of $\ce{PbSO4}$ is about $10^{-8}$. Now suppose $[\ce{Pb^{2+}}$] falls down from $\pu{1 M}$ to an arbitrary low value like $\pu{10^{-8} M},$ due to addition of $\ce{SO4^{2-}}$ ions. In this case, Nernst's law can be applied, and the potential of the lead electrode falls from $E^\circ_\ce{Pb} = \pu{-0.13 V}$ down to
$$E_\ce{Pb} = \pu{-0.13 V} + \pu{0.0296 V}\times\log{10^{-8}} = \pu{-0.37 V}\tag{1}$$
As a consequence, the $\ce{Pb}$ potential decreases. But the cell potential increases and goes from the initial value (without sulfate)
$$E^\circ_\mathrm{cell, i} = \pu{0.34 V} - (\pu{-0.13 V}) = \pu{0.47 V}\tag{2}$$
to the final value (after precipitation of lead sulfate)
$$E_\mathrm{cell, f} = \pu{0.34 V} - (\pu{-0.37 V}) = \pu{0.71 V}\tag{3}$$