I need help understanding this stability analysis for the given finite difference approximation, I don't understand the middle part where if we assume $p^{2} >1 $ so $p>1$ how is this the same as $ p<-1$ , where did assumption come from? Also if $p<-1$ how does this make $ \left\lvert \xi _-\right\rvert>1$? can someone please look at the solution and explain? Thank you.
Addressing your first question, $p^2>1$ implies that either $p>1$ or $p<-1$, which makes sense, since in case of negative $p$ it can be written as $p = -1\!\cdot\! r,\ r\ge0$, and so $$ p^2 = \left(-r\right)^2 = r^2 >1 \implies r >1 \implies p = -r <-1 $$
Also, since for the case $p^2>1$ we have
$$ 1-p^2\sin^2\omega<0 \qquad \text{and}\qquad \left\lvert\xi_-\right\rvert = \big\lvert \,p\sin\omega -\sqrt{p^2\sin^2\omega - 1}\, \big\rvert $$
The first condition $1-p^2\sin^2\omega<0$ means that the expression under root sign is non-negative. If $p<-1$ and $-1\le\sin\omega\le1$ then the term $p\sin\omega$ can take values less than $-1$ (e.g. $\omega=\frac\pi2$) and so subtracting the (nonnegative) root term from $p\sin\omega$ will increase magnitude of the whole expression for certain values of $\omega$.
Example
$$ p=-2, \omega=\dfrac{\pi}{3} \Rightarrow p\sin\omega = -\dfrac{\sqrt{3}}{2} \Rightarrow \left\lvert\xi_-\right\rvert = \Big\lvert \,-\dfrac{\sqrt{3}}{2} - \sqrt{\dfrac{9}{4} - 1}\, \Big\rvert = \Big\lvert \,\dfrac{-\sqrt{3}-\sqrt{5}}{2}\, \Big\rvert \approx 1.984. $$