$\because$ Continuous at $x=2$

$$\begin{gathered} \Rightarrow p\left(2\right)+\sqrt{q}=2^2-5\left(2\right)+6 \\ \Rightarrow \sqrt{q}=-2p\dots \left(\mathrm{i}\right) \end{gathered}$$
Continuous at $x=3$

$$\begin{gathered} \Rightarrow a^2\left(9\right)+b\left(3\right)+1=3^2-5\left(3\right)+6 \\ \Rightarrow 9a^2+3b+1=0\dots \left(\mathrm{ii}\right) \end{gathered}$$
$f^{\text{'}}\left(x\right)=\left\{\begin{array}{cc}p& x\le 2\\ 2x-5& 2<x<3\\ 2a^{2}x+b& x\ge 3\end{array}\begin{array}{}\\ \end{array}\right.$
$\because$ Differentiable at $x=2$

$$\begin{gathered} \Rightarrow p=2\left(2\right)-5 \\ \Rightarrow p=-1\dots ..\left(\mathrm{iii}\right) \end{gathered}$$
$\because$ Differentiable at $x=3$

$$\begin{gathered} \Rightarrow 2a^2\left(3\right)+b=2\left(3\right)-5 \\ \Rightarrow 6a^2+b=1\dots \dots \left(\mathrm{iv}\right) \end{gathered}$$
$\therefore p=-1,q=4,a=\frac{2}{3},b=-\frac{5}{3}$ {from equations $\left(\mathrm{i}\right),\left(\mathrm{i}\mathrm{i}\right),\left(\mathrm{i}\mathrm{i}\mathrm{i}\right)$ and $\left(\mathrm{i}\mathrm{v}\right)$}
$\left|p\right|+\left|q\right|+\left|\frac{1}{a}\right|+\left|\frac{1}{b}\right|=1+4+\frac{3}{2}+\frac{3}{5}=\frac{71}{10}$