Sujal, there is no compound as K2 HD I4 but in case you were asking about K2HgI4 or Potassium mercuric iodide, I have provided the solution below:

Van't Hoff factor is given by the formula: 1+(n-1).(alpha), where alpha is the degree of disassociation and n is the number of molecules on the product side of the reaction.

In the given question, alpha = 40% or 0.4

Also, the reaction upon ionization of K2HgI4 in aqueous solution is:

K2HgI4 -> 2K(+) + HgI4(2-)

So the number of molecules on product side is 2+1=3. Hence n=3.

So van't Hoff factor is 1+(n-1).(alpha)

= 1+(3-1).(0.4)

= 1+2.(0.4)

= 1+0.8

= 1.8