Overall reaction under discussion is the free-radical chlorination of methane: $$CH_4 + Cl_2 \rightarrow CH_3Cl + HCl$$.

To test each statement we must compare the bond energies of the bonds broken with those of the bonds formed. Standard gas-phase bond dissociation energies (in kcal mol$$^{-1}$$) are taken as:

$$\begin{aligned} \text{C-H (in CH}_4) &:& 104 \\[2pt] \text{Cl-Cl} &:& 58 \\[2pt] \text{H-Cl} &:& 103 \\[2pt] \text{C-Cl} &:& 85 \end{aligned}$$

Step-wise energetics

1. Initiation
$$Cl_2 \;\longrightarrow\; 2\,^{\bullet}Cl$$
Only the Cl-Cl bond is broken.
$$\Delta H^\circ_{\text{init}} = +58\text{ kcal mol}^{-1}$$ (endothermic).
Therefore Option A, which claims an exothermic value of −58 kcal mol$$^{-1}$$, is incorrect.

2. Propagation - radical formation
$$^{\bullet}Cl + CH_4 \;\longrightarrow\; ^{\bullet}CH_3 + HCl$$
Bonds broken : one C-H (104)   Bonds formed : one H-Cl (103)
$$\Delta H^\circ = 104 - 103 = +1\text{ kcal mol}^{-1}$$ (slightly endothermic).
Thus Option B (exothermic, +27 kcal mol$$^{-1}$$) is wrong.

3. Propagation - product formation
$$^{\bullet}CH_3 + Cl_2 \;\longrightarrow\; CH_3Cl + ^{\bullet}Cl$$
Bonds broken : one Cl-Cl (58)   Bonds formed : one C-Cl (85)
$$\Delta H^\circ = 58 - 85 = -27\text{ kcal mol}^{-1}$$ (exothermic).
Hence Option C, which states this step is endothermic and +27 kcal mol$$^{-1}$$, is also wrong.

4. Overall reaction
Bonds broken : C-H (104) + Cl-Cl (58) = 162
Bonds formed : C-Cl (85) + H-Cl (103) = 188
$$\Delta H^\circ_{\text{overall}} = 162 - 188 = -26\text{ kcal mol}^{-1} \approx -25\text{ kcal mol}^{-1}$$
The process is exothermic by roughly 25 kcal mol$$^{-1}$$, matching Option D.

Therefore, the only correct statement is:
Option D which is: The reaction is exothermic with $$\triangle H^\circ = -25$$ kcal mol$$^{-1}$$.