Acidic ferric chloride solution on treatment with excess of potassium ferrocyanide gives a Prussian blue coloured colloidal species. It is:

We have an acidic solution of ferric chloride. In water it dissociates completely:

$$\mathrm{FeCl_3 \;\longrightarrow\; Fe^{3+} + 3\,Cl^-}$$

Now we add an excess of potassium ferrocyanide, whose formula is $$\mathrm{K_4[Fe(CN)_6]}$$. It, too, dissociates in aqueous medium:

$$\mathrm{K_4[Fe(CN)_6] \;\longrightarrow\; 4\,K^+ + [Fe(CN)_6]^{4-}}$$

In the mixture we therefore have the following free ions:

$$\mathrm{Fe^{3+}},\; [Fe(CN)_6]^{4-},\; K^+,\; Cl^-$$

The coloured species that appears is produced by the direct combination of the ferric ion $$\mathrm{Fe^{3+}}$$ with the ferrocyanide ion $$\mathrm{[Fe(CN)_6]^{4-}}$$. Because $$\mathrm{Fe^{3+}}$$ carries a charge of $$+3$$ and $$\mathrm{[Fe(CN)_6]^{4-}}$$ carries a charge of $$-4$$, their initial 1 : 1 combination gives an anion of net charge $$-1$$:

$$\mathrm{Fe^{3+} + [Fe(CN)_6]^{4-} \;\longrightarrow\; [Fe^{3+}[Fe^{2+}(CN)_6]]^-}$$

We may simply write this single-negative complex ion as $$\mathrm{[Fe[Fe(CN)_6]]^-}$$.

Because the solution contains an excess of $$\mathrm{K^+}$$ (four potassium ions came with every ferrocyanide ion), one of those potassium ions immediately neutralises the single negative charge:

$$\mathrm{K^+ + [Fe[Fe(CN)_6]]^- \;\longrightarrow\; KFe[Fe(CN)_6]}$$

The electrically neutral species $$\mathrm{KFe[Fe(CN)_6]}$$ precipitates out as an extremely fine colloidal solid with an intense blue colour known as Prussian blue. (The stoichiometry never reaches $$\mathrm{Fe_4[Fe(CN)_6]_3}$$ because the large excess of potassium ferrocyanide always supplies enough $$\mathrm{K^+}$$ to trap the product in the one-potassium form.)

Thus the Prussian blue coloured colloidal compound produced under the stated conditions is

$$\boxed{\mathrm{KFe[Fe(CN)_6]}}$$

Hence, the correct answer is Option 4.