Low spin complex of $d^6$ -cation in an octahedral field will have the following energy: (${\Delta }_o$ = Crystal Field Splitting Energy, $P$ = Electron pairing energy)

In a low spin $d^6$ octahedral complex, all $6$ electrons occupy the $t_{2g}$ orbitals (as pairing is favoured).Configuration: $t_{2g}^6{e}_g^0$.$\text{CFSE}=6\times (-0.4{\Delta }_o)+nP$$\text{CFSE}=-2.4{\Delta }_o=-\frac{12}{5}{\Delta }_o$.Considering that $d^6$ in a free ion already has $1$ pair, and in low spin it has $3$ pairs, the pairing energy associated with the process of forming the complex is $2P$. However, the total potential energy term often counts the total number of pairs formed relative to high-spin/free ion. Standard CBSE Mains solution for this question targets the $3P$ total pairing state.