The group is generated by the three elements $ \sigma=\begin{pmatrix}1 & 1\\\\0&1\end{pmatrix}, x=\begin{pmatrix}\alpha & 0\\\\0&1\end{pmatrix}, \text{ and }y=\begin{pmatrix}1 & 0\\\\0&\alpha\end{pmatrix}, $ where $\alpha$ is a generator of $(\mathbb{Z}/p\mathbb{Z})^\times$. To get a presentation, you need to determine how these three elements conjugate with each other. Of course, as you have already said, $x$ and $y$ commute, so you just need to compute $x\sigma x^{-1}$ and $y\sigma y^{-1}$. E.g. $ x\sigma x^{-1} = \begin{pmatrix}\alpha & 0\\\\0&1\end{pmatrix}\begin{pmatrix}1 & 1\\\\0&1\end{pmatrix}\begin{pmatrix}\alpha^{-1} & 0\\\\0&1\end{pmatrix}= \begin{pmatrix}1 & \alpha\\\\0&1\end{pmatrix}=\sigma^{a}, $ for any lift $a$ of $\alpha\in\mathbb{Z}/p\mathbb{Z}$ to $\mathbb{Z}$. Similarly, you will find that $y\sigma y^{-1}=\sigma^{b}$ where $b$ is a lift of $\alpha^{-1}$ to $\mathbb{Z}$. This gives you the presentation $ B=\langle x,y,\sigma|x^{p-1}=y^{p-1}=\sigma^p = 1,xy=yx,x\sigma x^{-1}=\sigma^a,y\sigma y^{-1}=\sigma^{b}\rangle $ The last relation can also be replaced by $y^{-1}\sigma y = \sigma^a$, if you don't want to introduce $b$.