An octahedral ion has seven d electrons…

Step 1. High spin. $t_{2g}^5 e_g^2$ has $\text{CFSE} = (-2.0 + 1.2) \times 215 = -0.8 \times 215 = -172\ \text{kJ/mol}$. It has no extra pairs.

Step 2. Low spin. $t_{2g}^6 e_g^1$ has $\text{CFSE} = (-2.4 + 0.6) \times 215 = -1.8 \times 215 = -387\ \text{kJ/mol}$. It has 3 pairs and the free ion has 2, so 1 extra pair costs $245\ \text{kJ/mol}$. The total is $-387 + 245 = -142\ \text{kJ/mol}$.

Step 3. The winner. The high spin total is lower by $30\ \text{kJ/mol}$. This agrees with $\Delta_o < P$.

Step 4. Unpaired electrons. The high spin ion has $t_{2g}^5 e_g^2$. The $t_{2g}^5$ part has 1 unpaired electron and $e_g^2$ has 2, so there are 3 in all.