One mole of magnesium in the vapour state absorbed 1200 kJ/mol energy. If the first and second ionisation energies of Mg are 750 kJ/mol and 1450 kJ/mol respectively, the final composition of the mixture is:

Energy absorbed in the ionization of 1 mole of Mg to Mg+(g) = 750 kJ
Energy left unused = 1200 kJ - 750 kJ = 450 kJ
Since the second ionization energy (1450 kJ/mol) is greater than the remaining energy (450 kJ), only the first ionization will occur completely.
Therefore, the final composition of the mixture will be predominantly Mg+ with a small amount of remaining Mg.
The percentage of Mg+ can be calculated by considering that 750 kJ is required to ionize 100% of Mg. Therefore:
Percentage of Mg+ = (750 kJ / 1200 kJ) * 100% = 62.5%
Percentage of Mg = 100% - 62.5% = 37.5%
The closest option to 62.5% is 69%. However, this calculation assumes that all the absorbed energy goes into ionization. In reality, some energy might be lost as heat or to other processes. The problem lacks sufficient information to accurately determine the exact composition of the mixture beyond recognizing that it will consist predominantly of Mg+.