The combustion of benzene (l) gives CO$$_2$$(g) and H$$_2$$O(l). Given that heat of combustion of benzene at constant volume is $$-3263.9$$ kJ mol$$^{-1}$$ at 25$$^\circ$$C; the heat of combustion (in kJ mol$$^{-1}$$) of benzene at constant pressure will be:
(R = 8.314 JK$$^{-1}$$ mol$$^{-1}$$)

We are told that the heat liberated at constant volume is $$\Delta U = q_v = -3263.9\ \text{kJ mol}^{-1}$$ for the combustion of one mole of liquid benzene at $$25^\circ \text{C}$$ (that is, at $$T = 298\ \text{K}$$). We have to find the heat of combustion at constant pressure, i.e. the enthalpy change $$\Delta H$$.

For any chemical reaction carried out at temperature $$T$$, the well-known thermodynamic relation connecting the two heat terms is first stated:

$$\boxed{\Delta H = \Delta U + \Delta n_g R T}$$

Here $$\Delta n_g$$ is the algebraic change in the number of moles of gaseous species during the reaction, $$R$$ is the universal gas constant and $$T$$ is the absolute temperature. Solids and liquids are not counted in $$\Delta n_g$$ because only gaseous expansion or contraction does $$P\Delta V$$ work.

So we first write the balanced combustion equation for one mole of benzene:

$$\mathrm{C_6H_6(l) + \dfrac{15}{2}\,O_2(g) \;\longrightarrow\; 6\,CO_2(g) + 3\,H_2O(l)}$$

Now we count the gaseous moles on each side:

Reactants: only $$O_2(g)$$ is gaseous, giving $$n_{g,\text{react}} = \dfrac{15}{2} = 7.5$$.

Products: only $$CO_2(g)$$ is gaseous, giving $$n_{g,\text{prod}} = 6$$.

Therefore,

$$\Delta n_g = n_{g,\text{prod}} - n_{g,\text{react}} = 6 - 7.5 = -1.5$$

Next we substitute each quantity in the enthalpy relation. The gas constant must be in kJ for uniform units:

$$R = 8.314\ \text{J K}^{-1}\text{mol}^{-1} = 0.008314\ \text{kJ K}^{-1}\text{mol}^{-1}$$

So,

$$\Delta n_g R T = (-1.5)\,(0.008314\ \text{kJ K}^{-1}\text{mol}^{-1})\,(298\ \text{K})$$

First multiply $$R$$ and $$T$$:

$$0.008314 \times 298 = 2.476\ \text{kJ mol}^{-1}$$

Then multiply by $$\Delta n_g$$:

$$\Delta n_g R T = (-1.5)\,(2.476) = -3.714\ \text{kJ mol}^{-1}$$

Finally, substitute into the main formula:

$$\Delta H = \Delta U + \Delta n_g R T = (-3263.9) + (-3.714)$$

$$\Delta H = -3267.614\ \text{kJ mol}^{-1}$$

Rounding to the significant figures used in the problem gives

$$\boxed{\Delta H \approx -3267.6\ \text{kJ mol}^{-1}}$$

Among the options provided, this value corresponds to Option A.

Hence, the correct answer is Option A.