Answer
$Pd$ has the same number of electrons and the same electron configuration as $Sn^{4+}$.
Work Step by Step
$Sn^{4+}$
A neutral $Sn$ atom has 50 electrons. Losing 4 electrons to become $Sn^{4+}$ cation, it now has 46 electrons. The neutral atom that also has 46 electrons is $Pd$.
The electron configuration of $Sn$ is $[Kr]4d^{10}5s^25p^2$. Now we need to remove 4 electrons so that $Sn$ would become $Sn^{4+}$. Again, we would remove the electrons of the orbitals with the highest principal quantum number first, which is 5 here. The electrons in orbitals $5s$ and $5p$ are also 4, so all these 4 electrons will be removed.
Then, the electron configuration of $Sn^{4+}$ is $[Kr]4d^{10}$, which is also the electron configuration of $Pd$.
