For four elements the ${\mathrm{IP}}_2$ curve is shown.  In the graph the elements represented by $A, B, C$ and $D$ are

The second ionisation potential is presented as IP₂ is equal to the second ionisation energy. 

Mathematically, writing this relation for Mg, Al, Si and P we have:

For Mg:

MgZ=12→1 electron lossMg+→1s22s22p63s1 IP1 Mg+→1 electron lossMg2+→1s22s22p6 IP2

For Al:

AlZ=13→1 electron lossAl+→1s22s22p63s2 IP1 Al+→1 electron lossAl2+→1s22s22p63s1 IP2

For Si:

SiZ=14→1 electron lossSi+→1s22s22p63s23p1 IP1 Si+→1 electron lossSi2+→1s22s22p63s2 IP2

For P:

PZ=15→1 electron lossP+→1s22s22p63s23p2 IP1 P+→1 electron lossP2+→1s22s22p63s2 3p1 IP2

So, the element A is Mg as its second ionisation energy is less due to its instability and Al is more stable after the second ionisation, so having a higher value of IP2. Similarly, Si would have a smaller value of second ionisation energy due to its instability and P would have a higher value for the same as it becomes stable.

Hence, the correct labelling would be A→Mg, B→Al, C→Si, and D→P, making option B the correct answer.