Is enthalpy of combustion the 'heat released' or 'enthalpy change'?
The enthalpy of combustion (aka heat of combustion) is defined as the enthalpy change in going from pure reactants (each at 1 bar pressure) to pure products (again, each at 1 bar pressure) at 298 K.

In this problem, it is first of all necessary to assume ideal gas behavior to proceed.  For ideal gases, the change in enthalpy in going from pure reactants each at 1 bar pressure to the starting gas mixture in the calorimeter, there is no change in enthalpy at constant temperature.  Then, application of the first law of thermodynamics to the mixture before and after the reaction leads to the conclusion that $\Delta U=Q$, since the reaction volume is constant.  Since the temperature change between the beginning and end of the combustion in the calorimeter is essentially zero, this means the the calorimeter has measured the internal energy change of the reaction.  As Buck Thorn has pointed out, the change in enthalpy for the gases in the calorimeter is given for an ideal gas by:
$$\Delta H=\Delta U+\Delta (PV)=\Delta U+RT\Delta n$$where $\Delta n$ is the stoichiometric change in the number of moles of gas.  He also correctly pointed out that there is nothing said in the problem statement about the stoichometry of the reaction or the change in the number of moles of gas.  So, as he indicated, the problem statement is flawed.  However, if we arbitrarily assume that the change in the number of moles of gas is zero, we would have to conclude that the change in enthalpy for the process in the calorimeter is equal to the change in internal energy.  Then, since, again, the products form an ideal gas mixture, there is no change in enthalpy in going from the final gas mixture in the calorimeter to the pure products at 1 bar pressure.  So, based on these assumptions, the enthalpy of combustion is equal to the internal energy change in the calorimeter (also Q) per mole of species being burned.