Among the given options, the compounds that show color due to d-d transitions are those that have partially filled d-orbitals when in the form of complex ions or compounds.

Let's examine each option:

Option A: $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$

Chromium in $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ exists in the +6 oxidation state as the dichromate ion ($\mathrm{Cr}_2\mathrm{O}_7^{2-}$). In this oxidation state, chromium does not have any electrons in the d-orbitals (it has a $\mathrm{[Ar]}\;3\mathrm{d}^0$ configuration) so there is no possibility for d-d transitions. Instead, the color is due to charge transfer transitions.

Option B: $\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}$

Copper(II) sulfate pentahydrate, $\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}$, contains copper in a +2 oxidation state. In this state, copper has a $\mathrm{[Ar]}\;3\mathrm{d}^9$ electron configuration, which means there is one unpaired electron in the d-orbitals. Due to the presence of this unpaired electron, d-d transitions can occur, and this is why $\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}$ is blue in color.

Option C: $\mathrm{KMnO}_4$

In potassium permanganate ($\mathrm{KMnO}_4$), manganese is in the +7 oxidation state, and it has a $\mathrm{[Ar]}\;3\mathrm{d}^0$ electron configuration. Like chromium in the +6 state, manganese in the +7 state does not have any d-electrons available for d-d transitions. The deep purple color of $\mathrm{KMnO}_4$ is also due to charge transfer transitions.

Option D: $\mathrm{K}_2 \mathrm{CrO}_4$

Potassium chromate ($\mathrm{K}_2 \mathrm{CrO}_4$) contains chromium in the +6 oxidation state as the chromate ion ($\mathrm{CrO}_4^{2-}$). Like dichromate, chromium does not have d-electrons and therefore cannot undergo d-d transitions in this compound. The color is due to charge transfer transitions.

So the only compound from the options listed that shows color due to d-d transitions is Option B: $\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}$.