1. Identify the reaction between potassium iodide and potassium ferricyanide

Potassium ferricyanide is $K_3[Fe(CN)_6]$, containing the complex ion $[Fe(CN)_6]^{3-}$.

On adding iodide ions, a reversible redox reaction occurs:

$2[Fe(CN)_6]^{3-} + 2I^- \rightleftharpoons 2[Fe(CN)_6]^{4-} + I_2$

In excess iodide, the iodine formed combines with iodide to give triiodide:

$I_2 + I^- \rightleftharpoons I_3^-$

So the overall reversible reaction becomes:

$2[Fe(CN)_6]^{3-} + 3I^- \rightleftharpoons 2[Fe(CN)_6]^{4-} + I_3^-$

Thus the complex/species $P$ formed in solution is essentially the ferrocyanide-containing system, i.e. $[Fe(CN)_6]^{4-}$ produced in acidic medium as the equilibrium shifts to the right.

2. Why strong acid shifts equilibrium toward $P$

In strong acidic medium, the iodine/triiodide side is favored and the redox process proceeds essentially completely, generating ferrocyanide ion $[Fe(CN)_6]^{4-}$.

So for the precipitation step, the relevant anion is:

$[Fe(CN)_6]^{4-}$

3. Reaction with zinc chloride in slightly acidic medium

Zinc ion gives a sparingly soluble zinc ferrocyanide with ferrocyanide ion. The known insoluble zinc ferrocyanide is:

$Zn_2[Fe(CN)_6]$

because charge balance requires:

$2 \times (+2) + (-4) = 0$

Hence the precipitate $Q$ is zinc ferrocyanide, containing 2 zinc ions per formula unit.

4. Answer

The number of zinc ions in the molecular formula of $Q$ is:

$\boxed{2}$

5. Comparison with stored answer

The stored correct answer is given as 3 OR 2. My derived answer is 2, which matches one of the accepted answers.