Predict the species formed by the reaction of mercury(II) chloride and iodine
You are right about the formation of $\ce{I3-}$ ion (from the reaction of $\ce{I2}$ and solution containing iodide ion, let's say $\ce{KI}$ solution). Reaction between $\ce{HgCl2}$ and $\ce{KI}$ solution at $\pu{70 ^\circ C}$ produces mercury(II) iodide ($\ce{HgI2}$):
$$\ce{HgCl2 + 2 KI → HgI2 + KCl}$$
But the story doesn't end here. This mercury(II) iodide reacts with $\ce{KI}$ solution to form potassium tetraiodomercurate(II)($\ce{K2HgI4}$):

Crystallizing from a concentrated aqueous solution of mercuric iodide
with potassium iodide is the monohydrate $\ce{KHgI3.H2O}$, which is pale
orange. In aqueous solution, this triodido complex adds iodide to
give the tetrahedral tetraiodo dianion.

So, the answer is  C. but there is a typo. It should be $\ce{[HgI4]^2-}$. The overall reaction is:
$$\ce{HgCl2 + 5KI + I2 -&gt; K2HgI4 + KI3 + 2KCl}$$