In the exercise 5.8 Bump has asked to prove that the group $Sp(4)$ over complex numbers, which is usual complex embedding $U(4)\cap Sp(4,\mathbb{C})$, can be described by, $$\left\{\begin{pmatrix} a&b\\-\bar{b}&\bar{a} \end{pmatrix}: a,b\in\mathbb{H},|a|^2+|b|^2=1\right\}.$$ But how can that be true while we know that $Sp(4)\cong SO(5)$ is a $10$ dimensional group over $\mathbb{R}$ and the above is $7$ dimensional?
If the above is not true, how can we paramatrize $Sp(4)$?
I do not think $Sp_{4}(\mathbb{R})$ and $SO_{5}(\mathbb{R})$ are isomorphic. In particular I think their root systems are different by checking the classification diagram. Another way to see it is the fundamental group for $SO_{5}(\mathbb{R})$ is $\mathbb{Z}/2\mathbb{Z}$, while for $Sp_{4}(\mathbb{R})$ it is trivial.
For your question, I think David Bump wanted to show that it is equivalent to quaterionic matrices which preserve the inner product. This force its determinant to be 1 and $AA^{T}=I$. Therefore if we let the matrix to be $$ \begin{pmatrix} a&b\\ c&d\\ \end{pmatrix},a,b,c,d\in \mathbb{H} $$ We would have $|a|^{2}+|b|^{2}=1, |c|^{2}+|d|^{2}=1, ad-bc=1$. Together we get $1+1+4=6$ equations as the last one projects into coordinates. Assuming non-degenercy we get dimension $10=16-6$ as desired. But I do not know how to reduce this to David Bump's formula above, or if there is a nice form in general.