All the combinations are allowed for a $p^2$ configuration. This produces term symbols $^1\mathrm D$, $^3\mathrm P$ and $^1\mathrm D$, but you are asked for the ground state which by Hundt's rules is the $^3\mathrm P$.

This has $M_l= m_{l1} + m_{l2} = 1,0,-1$ and spin $M_s= m_{s1}+m_{s2}= 1,0,-1$ which makes 9 combinations.

The $^1\mathrm D$ and $^1\mathrm S$ must have $m_{s1}+m_{s2}= 0$ but S has $m_{l1}+m_{l2}= 0$ and so one combination and D has $m_{l1} + m_{l2} = 2,1,0,-1,2$ making 5 combinations as $M_s=0$ only. (The spin multiplicity superscript on the term symbol is $2M_s+1$).

The first entry has $M_l=1, M_s=0 $ so belongs to the singlet state $^1\mathrm D$.

The second entry has $M_l=0, M_s = 1$ so $\rightarrow ^3\mathrm P$.

The third entry has $M_l = 2,M_s=0$ so $\rightarrow ^1\mathrm D$

The forth entry has $M_l = 0,M_s=-1$ so $\rightarrow ^3\mathrm P$.

All the combinations listed are allowed ones for a $p^2$ configuration. (There are 15 combinations in total when Pauli exclusion is added). These combinations produce term symbols $^1\mathrm D$, $^3\mathrm P$ and $^1\mathrm D$, however, you are asked for the ground state which by Hundt's rules is the $^3\mathrm P$.

This has $M_l= m_{l1} + m_{l2} = 1,0,-1$ and spin $M_s= m_{s1}+m_{s2}= 1,0,-1$ which makes 9 combinations.

The $^1\mathrm D$ and $^1\mathrm S$ must have $m_{s1}+m_{s2}= 0$ but S has $m_{l1}+m_{l2}= 0$ and so one combination and D has $m_{l1} + m_{l2} = 2,1,0,-1,2$ making 5 combinations as $M_s=0$ only. (The spin multiplicity superscript on the term symbol is $2M_s+1$).

The first entry in your table has $M_l=1, M_s=0 $ so belongs to the singlet state $^1\mathrm D$.

The second entry has $M_l=0, M_s = 1$ so $\rightarrow ^3\mathrm P$.

The third entry has $M_l = 2,M_s=0$ so $\rightarrow ^1\mathrm D$

The forth entry has $M_l = 0,M_s=-1$ so $\rightarrow ^3\mathrm P$.