Chọn phương án B.

Ta có \(\log_25\cdot\log_53=\log_23\).
Suy ra \(\dfrac{\log_23}{\log_53}=\log_25\). Khi đó: $$\begin{aligned}
\dfrac{a(m+nb)}{b(a+p)}&=\dfrac{\log_23\left(m+n\log_53\right)}{\log_53\left(\log_23+p\right)}\\
\Leftrightarrow\log_645&=\dfrac{\log_25\left(m+n\log_53\right)}{\log_23+p}\\
\Leftrightarrow\log_645&=\dfrac{m\log_25+n\log_25\cdot\log_53}{\log_23+p}\\
\Leftrightarrow\log_645&=\dfrac{m\log_25+n\log_23}{\log_23+p}\\
\Leftrightarrow\log_645&=\dfrac{\log_25^m+\log_23^n}{\log_23+\log_22^p}\\
\Leftrightarrow\log_645&=\dfrac{\log_2\left(5^m\cdot3^n\right)}{\log_2\left(3\cdot2^p\right)}\\
\Leftrightarrow\log_645&=\log_{\left(3\cdot2^p\right)}\left(5^m\cdot3^n\right)
\end{aligned}$$
Theo đó $$\begin{cases}
3\cdot2^p&=6=3\cdot2^1\\
5^m\cdot3^n&=45=5^1\cdot3^2
\end{cases}\Rightarrow\begin{cases}
p=1\\ m=1\\ n=2
\end{cases}$$
Vậy \(m+n+p=1+2+1=4\).