Since, $a+b=-p, a b=1$ .....(i)
and $c+d=-q, c d=1$
Now $(a-c)(b-c)$ and $(a+d)(b+d)$ are the roots of $x^2+a x+\beta=0$
$$
\begin{aligned}
& (a-c)(b-c)(a+d)(b+d)=\beta \\
& \Rightarrow\left(a b-a c-b c+c^2\right)\left(a b+a d+b d+d^2\right)=\beta \\
& \Rightarrow\left\{1-c(a+b)+c^2\right\}\left\{1+d(a+b)+d^2\right\}=\beta \\
& \Rightarrow \quad\left(1+p c+c^2\right)\left(1-p d+d^2\right)=\beta \\
& \Rightarrow 1-p d+d^2+p c-p^2 c d+p c d^2+c^2-p c^2 d \\
& +c^2 d^2=\beta \\
& \Rightarrow 1-p d+d^2+p c-p^2+p d+c^2-p c+1=\beta \\
& {[\because c d=1]} \\
& \Rightarrow \quad 2+d^2+c^2-p^2=\beta \\
&
\end{aligned}
$$
$\begin{aligned} \Rightarrow \quad 2 c d+c^2+d^2-p^2 & =\beta \\ (c+d)^2-p^2 & =\beta \\ q^2-p^2 & =\beta \quad[\because(c+d)=-q]\end{aligned}$