Given region,

$$ E=\left\{(x, y): 3-x \leq y \leq \sqrt{9-x^2}, 0 \leq x \leq 3\right\} $$

Since, point $(p, p+1)$ lie on line $y=x+1$

$\therefore$ Point of intersection of $y=x+1$ and $y=3-x$

i.e., $x+1=3-x$

$\Rightarrow$ $2 x=2 \Rightarrow x=1$

and $y=2$

and point of intersection of $y=x+1$ and

$$ y=\sqrt{9-x^2} $$

i.e., $(x+1)^2=9-x^2$

$\begin{array}{lc} &\Rightarrow x^2+1+2 x=9-x^2 \\\\ &\Rightarrow 2 x^2+2 x-8=0 \\\\ &\Rightarrow x^2+x-4=0 \\\\ &\Rightarrow x=\frac{-1 \pm \sqrt{1+4(1)(4)}}{2} \\\\ &\Rightarrow x=\frac{-1 \pm \sqrt{17}}{2}\end{array}$

$\begin{array}{lll}\Rightarrow & x=\frac{-1+\sqrt{17}}{2}, & \text { (Since, } x \in[0,3]) \\\\ \therefore & p \in \left(1, \frac{-1+\sqrt{17}}{2}\right)\end{array}$

$\Rightarrow a=1, b=\frac{-1+\sqrt{17}}{2}$

$\begin{aligned} & \therefore b^2+b-a \\\\ &= \frac{1+17-2 \sqrt{17}}{4}+\frac{(-1+\sqrt{17})}{2}-1 \\\\ &= \frac{18-2 \sqrt{17}-2+2 \sqrt{17}-4}{4} \\\\ &= \frac{12}{4}=3\end{aligned}$