Answer : The correct option is, the sum of the enthalpy changes of the intermediate reactions.
Explanation :
According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.
For example :
The formation reaction of CO will be,

The intermediate balanced chemical reaction will be,
(1)

(2)

The expression for enthalpy of formation of CO is,
![\Delta H_(formation)=[n* \Delta H_1]+[n* \Delta H_2]](https://img.qammunity.org/2018/formulas/chemistry/high-school/8blem6kpqurv2ab3g617efhgj3uvvzbqfc.png)
where,
n = number of moles
By adding reaction 1 and reverse reaction of reaction 2, we get the enthalpy of formation of CO.
Hence, the correct option is, the sum of the enthalpy changes of the intermediate reactions.