Step 1: Recall the trend in Group 14 tetrahalides.
Group 14 elements form tetrahalides in the $+4$ state, but the stability of $+4$ drops down the group due to the inert pair effect, which makes lead prefer $+2$.
Step 2: Focus on lead.
Because $\mathrm{Pb^{2+}}$ is more stable than $\mathrm{Pb^{4+}}$, lead tetrahalides are inherently fragile in the $+4$ state.
Step 3: Bring in the role of the halide.
A tetrahalide also fails if the halide can reduce the metal. Iodide is a strong reducing agent, so it cannot survive next to the strongly oxidising $\mathrm{Pb^{4+}}$.
Step 4: Show the redox conflict for $PbI_4$.
$\mathrm{Pb^{4+}}$ oxidises $\mathrm{I^-}$ to $I_2$ while itself being reduced to $\mathrm{Pb^{2+}}$, so $PbI_4$ decomposes: $\mathrm{PbI_4 \rightarrow PbI_2 + I_2}$.
Step 5: Check the others survive.
$CCl_4$ and $SiCl_4$ are stable, and $PbCl_4$ exists under controlled cold conditions because chloride is a weaker reducing agent than iodide.
Step 6: Conclude.
Hence the tetrahalide that does not exist is $PbI_4$, which is option (4). \[ \boxed{\mathrm{PbI_4}} \]