The correct order of energy of absorption for the following metal complexes is
$$A: [Ni(en)_3]^{2+}$$, $$B: [Ni(NH_3)_6]^{2+}$$, $$C: [Ni(H_2O)_6]^{2+}$$

To determine the correct order of absorption energies for the nickel(II) complexes with different ligands, we use the spectrochemical series, which arranges ligands by increasing crystal field splitting energy ($$\Delta$$):

$$H_2O < NH_3 < en \text{ (ethylenediamine)}$$

A stronger field ligand produces a larger crystal field splitting energy $$\Delta_o$$, so the complex absorbs light of higher energy (shorter wavelength). The energy of absorption is given by:

$$E_{absorbed} = \Delta_o$$

Applying the ligand order to this relation yields:

$$[Ni(H_2O)_6]^{2+} < [Ni(NH_3)_6]^{2+} < [Ni(en)_3]^{2+}$$

Accordingly, if C, B, and A represent [Ni(H_2O)_6]^{2+}, [Ni(NH_3)_6]^{2+}, and [Ni(en)_3]^{2+} respectively, then:

$$C < B < A$$

Therefore, the correct choice is Option A.