From § 8.1, we have \begin{equation} \Sp(4,\RR) = \{ M\in\RR^{4\times 4} : M\Omega M^T = \Omega \} \end{equation} where \begin{equation} \Omega = \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \end{pmatrix} \end{equation}

Consider now the antisymmetric matrix \begin{equation} P = \begin{pmatrix} 0 & p+q & -x & t+z \\ -p-q & 0 & z-t & x \\ x & t-z & 0 & p-q \\ -t-z & -x & q-p & 0 \end{pmatrix} \end{equation} and note that \begin{equation} P^T\Omega P = P\Omega P^T = |P|^2\Omega \end{equation} where \begin{equation} |P|^2 = -t^2+x^2+z^2+p^2-q^2 \end{equation} (and is also the square root of $\det P$). We claim that $M\in\Sp(4,\RR)$ acts on $P$ via \begin{equation} P \longmapsto M^T P M \label{symtran2} \end{equation} that is, we claim that $M^T P M$ is a matrix of the same form as $P$. What form is that? $P$ is clearly antisymmetric, and the transformation (\ref{symtran2}) clearly takes antisymmetric matrices to antisymmetric matrices, since \begin{equation} (M^T P M)^T = M^T P^T M = - M^T P M \end{equation} However, antisymmetric matrices in 4 (real) dimensions contain six independent components, and we only have five. Which degree of freedom are we missing? That's easy: $P$ does not contain a multiple of $\Omega$ itself. But that's fine, since (\ref{symtran2}) takes $\Omega$ to itself, so if there's no $\Omega$ “component” to start with, then there won't be one afterward.

So what does the transformation (\ref{symtran2}) do? We have \begin{align} (M^T P M) \Omega (M^T P M)^T &= M^T P (M\Omega M^T) P^T M \nonumber\\ &= M^T (P \Omega P^T) M = M^T |P|^2 \Omega M \nonumber\\ &= |P|^2 \Omega \end{align} since $|P|^2$ is a real number, and therefore commutes with $M^T$. But $|P|^2$ is just the norm on $\RR^5$ with signature $(3,2)$, which is thus preserved by $\Sp(4,\RR)$. Since both $\Sp(4,\RR)$ and $\SO(3,2)$ are 10-dimensional, we have shown that \begin{equation} \Sp(4,\RR) \cong \Spin(3,2) \end{equation} where $\Spin(3,2)$ is of course the double cover of $\SO(3,2)$.