When potassium iodide and copper chloride combine, the white precipitate is : 

When two soluble ionic compounds react to produce a precipitate by exchanging their ions, such reactions are precipitation reactions.

One such reaction is: 

Hydrochloric acid and copper carbonate combine to generate copper (II) chloride and carbonic acid, which break down into carbon dioxide and water very quickly. (It breaks down almost immediately).

Due to the copper (II) ions present, this solution turns blue.

${\mathrm{CuCO}}_3+2\mathrm{HCl}\to {\mathrm{CuCl}}_2+{\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O}$

When potassium iodide and copper chloride combine, potassium chloride and copper iodide are both produced. ${\mathrm{Cu}}_2{\mathrm{I}}_2$ appears as a white precipitate.

$\begin{array}{rcl}{\mathrm{CuCl}}_2+2\mathrm{KI}& \to & {\mathrm{CuI}}_2+2\mathrm{KCl}\end{array}$

$\begin{array}{rcl}2{\mathrm{CuI}}_2& \to & {\mathrm{Cu}}_2{\mathrm{I}}_2+{\mathrm{I}}_{2 }\end{array}$

Therefore, the white precipitate is ${\mathrm{Cu}}_2{\mathrm{I}}_2$

So, option 1) is correct.