$$\ce{A_n ->[$k$] nA}$$
For the reaction given above, rate of formation of $\ce{A}$ is defined as:
$$\frac{1}{n}\frac{\mathrm d\ce{[A]}}{\mathrm dt} = k\ce{[A_n]}$$
In this case, we have three different reactions that involve A. They are:
\begin{align}
\ce{A_n &->[$k_1$] nA} \tag{1}\label{1}\\
\ce{nA &->[$k_1'$] A_n} \tag{2}\label{2}\\
\ce{A + B &->[$k_2$] P} \tag{3}\label{3}\\
\end{align}
Writing the rate equations for $\eqref{1}$, $\eqref{2}$, $\eqref{3}$ and summing them up to get total rate, we get:
\begin{align}
\frac{1}{n}\frac{\mathrm d\ce{[A]1}}{\mathrm dt} &= k_1\ce{[A_n]} \tag{4}\label{4}\\
-\frac{1}{n}\frac{\mathrm d\ce{[A]2}}{\mathrm dt} &= k'_1 [\ce{A}]^n\tag{5}\label{5}\\
-\frac{\mathrm d\ce{[A]3}}{\mathrm dt} &= k_2\ce{[A][B]} \tag{6}\label{6}\\
\frac{\mathrm d\ce{[A]_\mathrm{tot}}}{\mathrm dt} &= \frac{\mathrm d\ce{[A]1}}{\mathrm dt} + \frac{\mathrm d\ce{[A]2}}{\mathrm dt}+\frac{\mathrm d\ce{[A]3}}{\mathrm dt} \tag{7}\label{7} \\
\end{align}
Substituting values from $\eqref{4}$, $\eqref{5}$, $\eqref{6}$ into $\eqref{7}$, we get:
$$\frac{\mathrm d\ce{[A]_\mathrm{tot}}}{\mathrm dt} = \color{red}{n}k_1\ce{[A_n]} - \color{red}{n}k'_1\ce{[A]^n} - k_2\ce{[A][B]} $$
Here, $n = 2$, therefore
$$\frac{\mathrm d\ce{[A]_\mathrm{tot}}}{\mathrm dt} = \color{red}{2}k_1\ce{[A_n]} - \color{red}{2}k'_1\ce{[A]^2} - k_2\ce{[A][B]} $$
