Step 1: Find the metal's oxidation state and d-count.
In $[MnX_6]^{4-}$ each ligand is $-1$, so $x + 6(-1) = -4$ gives $x = +2$. So we have $Mn^{2+}$, which is $[Ar]\,3d^5$. Both complexes are therefore $d^5$ systems, and the only difference is the strength of the ligand field.
Step 2: Decide high-spin versus low-spin.
$X^-$ is a strong field ligand, so its splitting $\Delta_{o1}$ is large and beats the pairing energy $P$. Electrons pair up, giving the low-spin arrangement $t_{2g}^{5}e_g^{0}$. $Y^-$ is weak field, so $\Delta_{o2}$ is small, electrons stay unpaired, giving the high-spin arrangement $t_{2g}^{3}e_g^{2}$.
Step 3: Test statement (A).
Pairing only happens in the strong field complex because the field energy is big enough to force it. That is exactly the condition $\Delta_{o1} > P$, i.e. the pairing energy is smaller than $\Delta_{o1}$. So statement (A) is correct.
Step 4: Test statement (B) by comparing CFSE.
For low-spin $d^5$: $\text{CFSE} = 5(-0.4\,\Delta_{o1}) = -2\,\Delta_{o1}$. For high-spin $d^5$: $\text{CFSE} = 3(-0.4\,\Delta_{o2}) + 2(+0.6\,\Delta_{o2}) = 0$. The strong field complex is far more stabilised, so $[MnY_6]^{4-}$ is NOT more stable than $[MnX_6]^{4-}$. Statement (B) is wrong.
Step 5: Test statement (C).
A single $t_{2g}$ electron is lowered by only $0.4\,\Delta_o$ relative to the unsplit d-orbitals, not by $2\,\Delta_{o1}$. The figure $2\,\Delta_{o1}$ is the total CFSE for all five electrons together, not the per-orbital stabilisation claimed. So statement (C) is wrong.
Step 6: Test statement (D) and conclude.
The high-spin weak field complex has more unpaired electrons and more spin-allowed d-d transitions, so $[MnY_6]^{4-}$ absorbs more strongly and looks more intensely coloured. Statement (D) is correct. Collecting the verdicts, (A) and (D) are the correct statements.
\[ \boxed{(A)\ \text{and}\ (D)} \]