Answer
we see that the products of the carbon cycle mentioned are $2e^{+} + 2\nu + ^{4}He$ which is same as the proton-proton cycle as mentioned in the textbook. The difference is not significant and therefore, these reactions can be considered similar.
Work Step by Step
The carbon cycle given in the question is as follows
$\begin{array}{rll}^{12} \mathrm{C}+^{1} \mathrm{H} \rightarrow^{13} \mathrm{N}+\gamma, & Q_{1} & =1.95 \mathrm{MeV} \\ ^{13} \mathrm{N} \rightarrow^{13} \mathrm{C}+\mathrm{e}^{+}+\nu, & Q_{2} & =1.19 \\ ^{13} \mathrm{C}+^{1} \mathrm{H} \rightarrow^{14} \mathrm{N}+\gamma, & Q_{3} & =7.55 \\ ^{14} \mathrm{N}+^{1} \mathrm{H} \rightarrow^{15} \mathrm{O}+\gamma, & Q_{4} & =7.30 \\ ^{15} \mathrm{O} \rightarrow^{15} \mathrm{N}+\mathrm{e}^{+}+\nu, & Q_{5} & =1.73 \\ ^{15} \mathrm{N}+^{1} \mathrm{H} \rightarrow^{12} \mathrm{C}+^{4} \mathrm{He}, & Q_{6} & =4.97\end{array}$
If we add all the equations, to see the resulting equation, we see that the products of the carbon cycle mentioned are $2e^{+} + 2\nu + ^{4}He$ which is same as the proton-proton cycle as mentioned in the textbook. The difference is not significant and therefore, these reactions can be considered similar.
