How to calculate an enthalpy change for this reaction
To preface, the reason that this is not a combustion reaction is because this is a redox reaction that occurs at room temperature. In fact, chloroform is noncombustible.

Though at first glance, this may simply appear to be an unbalanced equation that could be rewritten: 
$$\ce{CHCl3 + 1/2O2 -&gt; COCl2 + HCl}$$

It is better to take it for what the question presents it as, an oxygen atom, so I would therefore recommend you to leave the equation as:

$$\ce{CHCl3 + O -&gt; COCl2 + HCl}$$

Though $\ce{O2}$ is a pure element, and its $\Delta H^0_\mathrm f=0\:\mathrm{kJ\:mol^{-1}}$, for the atom $\ce{O}$, $\Delta H^0_\mathrm f=58.99\:\mathrm{kJ\:mol^{-1}}^{[1]}$ (provided thanks to user1420303).

To calculate $\Delta H$, the equation is:  $$\Delta H=\Delta H^0_\mathrm f(\ce{COCl2}) + \Delta H^0_\mathrm f(\ce{HCl})-\Delta H^0_\mathrm f(\ce{CHCl3})-\Delta H^0_\mathrm f(\ce{O})$$

Though enthalpy is a state function, this question is asking you to calculate $\Delta H$ from a specific set of reactants, and their individual energies must be considered. There is more than just one way to calculate $\Delta H$, and an analysis of the $\Delta H^0_\mathrm f$ (the enthalpy of formation of a compound from its constituent elements) for each compound accounts for which bonds are being broken and formed, and will yield (about) the same results as a bond by bond analysis. I find this method to be preferable because it bond by bond analyses are based on average bond enthalpies, while $\Delta H^0_\mathrm f$ is often painstakingly measured for each compound, and will yield the most accurate $\Delta H$ calculation.



[1] Curtiss, L. A.; Raghavachari, K.; Redfern, P. C.; Pople, J. A. Assessment Of Gaussian-2 and Density Functional Theories for the Computation of Enthalpies of Formation. The Journal of Chemical Physics J. Chem. Phys. 1997, 106, 1063.