${\mathrm{Fe}}^{2+}$ does not give prussian blue colour with ${\mathrm{K}}_4\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]$ but on its reaction with $\left(\mathrm{X}\right)$, prussian blue colour appears $\left(\mathrm{X}\right)$ can be:

Prussian blue is produced by combining  ${\mathrm{Fe}}^{2+}$ with ${\mathrm{MnO}}_4^{-}/{\mathrm{H}}^{+}$. As oxidising agents, they change ferrous metal into ferric ions. This ferric ion interacts with ${\mathrm{K}}_4\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]$ to produce ferric Ferro cyanide, which has the colour of Prussian blue.

$\begin{array}{l}{\mathrm{MnO}}_4/{\mathrm{H}}^{+}\to \left[\mathrm{O}\right];{\mathrm{Fe}}^{2+}\to {\mathrm{Fe}}^{3+}\\ {\mathrm{Fe}}^{3+}+{\mathrm{K}}_4\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]\to {\mathrm{Fe}}_4{\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]}_3+4{\mathrm{K}}^{+}\end{array}$

Hence, ${\mathrm{MnO}}_4^{-}/{\mathrm{H}}^{+}$ does not give Prussian blue colour with ${\mathrm{K}}_4\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]$.