How does a limiting reactant relate to the concentration of ions in solution?
Of course, the concentration of $\ce{Na^+}$ is $0.2\ce{M}$, because it is a spectator ion. It doesn't participate in the  main reaction between  $\ce{Pb^2+}$ and $\ce{SO4^2-}$.
The concentration of $\ce{Na^+}$ is $0.1 \times 2=0.2 \ce{M}$ as there are two ions of sodium in sodium sulfate.

The number of mols  of ion nitrate, as a spectator ion, is $2\times1.5/331.2= 0.00906 \ce{mol}$.
The concentration of ion nitrate is  $ 0.00906/0.125=0.0725 \ce{M}$.

Let's go back to the main reaction between  $\ce{Pb^2+}$ and $\ce{SO4^2-}$: The limiting ion is $\ce{Pb^2+}$ as the number of moles of lead nitrate  $1.5/331.2=0.00453 \ce{mol}$ is smaller than the number of moles of sodium sulfate  $0.0125 \ce{mol}$.
So, after the reaction, there  are no  $\ce{Pb^2+}$ ions left in solution.

As for the remaining $\ce{SO4^2-}$; its concentration is:  $(0.0125-0.00453)/0.125=0.06376 \ce{M}$