1. Identify the reversible reaction between potassium iodide and potassium ferricyanide

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

Iodide ion $I^-$ acts as a reducing agent and reduces ferricyanide to ferrocyanide:

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

In terms of potassium salts:

$2K_3[Fe(CN)_6] + 2KI \rightleftharpoons 2K_4[Fe(CN)_6] + I_2$

So the complex $\mathbf{P}$ formed is:

$\mathbf{P} = K_4[Fe(CN)_6]$

2. Why does strong acidic medium shift equilibrium towards $P$?

In strong acid, the iodine formed is consumed by iodide to give species like $I_3^-$/further side processes, and effectively the reaction is driven forward. The important identification remains that $P$ is potassium ferrocyanide, $K_4[Fe(CN)_6]$.

3. Identify $Q$ formed with zinc chloride in slightly acidic medium

Ferrocyanide gives a sparingly soluble zinc ferrocyanide with $Zn^{2+}$:

$2Zn^{2+} + [Fe(CN)_6]^{4-} \rightarrow Zn_2[Fe(CN)_6] \downarrow$

Thus this confirms again that $P$ contains the ferrocyanide ion.

4. Find moles of KI required to produce 2 moles of $P$

From the balanced reaction:

$2K_3[Fe(CN)_6] + 2KI \rightarrow 2K_4[Fe(CN)_6] + I_2$

This shows:

$2\text{ mol } KI \rightarrow 2\text{ mol } P$

Therefore,

$1\text{ mol } KI \rightarrow 1\text{ mol } P$

Hence, to produce $2$ moles of $P$, required moles of $KI$ are:

$\boxed{2}$

5. Comparison with stored answer

Derived answer = $2$

Stored correct answer = $2$

So they agree.