Step 1: Rule of oxidation state:
For the same metal, the higher oxidation state gives the more covalent halide. Lead appears as $\text{Pb}^{2+}$ and $\text{Pb}^{4+}$.
Step 2: Compare:
$\text{PbCl}_4$ is the +4 chloride of lead, so it is more covalent than $\text{PbCl}_2$ (D).
The cations in $\text{SnCl}_2$ (A) and $\text{SbCl}_3$ (C) carry a lower charge (+2 and +3) than +4, so their polarising power is lower than that of $\text{Pb}^{4+}$.
Final Answer:
The most covalent halide is $\text{PbCl}_4$, option (B).
\[ \boxed{\text{PbCl}_4} \]