Hydrated copper sulphate $\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}$ is blue in colour. The ligand (water) molecules causes splitting of $d$-orbitals. This facilitated $d$ - $d$ transition and colour.
Anhydrous copper sulphate $\mathrm{CuSO}_4$ is colourless. In the absence of ligand (water) molecules, splitting of $d$-orbitals is not possible. Hence, $d$ - $d$ transition is not possible. Hence, option (b) is correct.
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