Step 1: Recall what $t_{2g}$ and $e_g$ represent.
In an octahedral field the five d-orbitals split into a lower triplet, $t_{2g}$, and a higher doublet, $e_g$, separated by the crystal field splitting energy $\Delta_0$, while $P$ stands for the pairing energy needed to force two electrons into one orbital.
Step 2: Interpret the condition $\Delta_0 < P$.
This inequality tells us the splitting gap is small, which is the signature of a weak field ligand, so it costs an electron less energy to jump up to $e_g$ than to pair up with another electron already sitting in $t_{2g}$.
Step 3: Fill the four electrons one at a time following Hund's rule.
The first three electrons go singly into the three $t_{2g}$ orbitals, giving $t_{2g}^3$, and since pairing is the costlier option here, the fourth electron avoids $t_{2g}$ and instead occupies one of the empty $e_g$ orbitals.
Step 4: Write the final configuration.
This gives the high-spin arrangement expected under a weak field, with the maximum possible number of unpaired electrons for a $d^4$ ion. \[ \boxed{t_{2g}^3\, e_g^1} \]