Let $\mathbf G$ be a connected, reductive group over a local field $F$. Assume there is a maximal torus $\mathbf T$ which is split over $F$. Let $\mathbf P$ be a parabolic subgroup of $\mathbf G$ containing a Borel subgroup $\mathbf B \supseteq \mathbf T$, and let $\mathbf P = \mathbf M \mathbf N$ be a Levi decomposition. Let $G = \mathbf G(F), P = \mathbf P(F)$ etc.
Here's a nice way to calculate the Haar measure on $P$. Let $\mu$ be a Haar measure on $N$ which, being nilpotent, is unimodular. Since $P$ is the semidirect product of $M$ and $N$, it suffices to compute how the measure on $N$ changes under conjugation by $M$. That is, there exists a homomorphism $\delta: M \rightarrow (0,\infty)$ such that $\mu(E) = \delta(m) \mu(mEm^{-1})$ for all $E \subseteq N$ Borel and all $m \in M$.
As topological spaces,
$$N = \prod\limits_{\alpha \in \Phi(T,N)} U_{\alpha}$$
where $U_{\alpha}$ is isomorphic to $F$ via an isomorphism $x_{\alpha}: F \rightarrow U_{\alpha}$ such that $$x_{\alpha}(\alpha(t)a) = t x_{\alpha}(a)t^{-1}$$ for all $t \in T$, $a \in F$. Choose an absolute value $| \cdot |$ on $F$ and a Haar measure $\mu_F$ on $F$ such that $\mu_F(aC) = |a| \mu_F(C)$ for all $C \subseteq F$ measurable and $a \in F^{\ast}$.
Now $\mathbf M$ is generated by its radical $\mathbf A$ and its derived group. Since $\delta$ is trivial on the derived group, to calculate $\delta(m)$, it suffices to calculate $\delta(a)$ for $a \in A$. Choose various$E_{\alpha} \subseteq F$ with nonzero finite measure, and let $$E = \prod\limits_{\alpha \in \Phi(T,N)} x_{\alpha}(E_{\alpha})$$
Using the fact that the Haar measure on $N$ is the product of the Haar measures on $U_{\alpha}$, we have
$$\mu(aEa^{-1}) = \prod\limits_{\alpha} \mu_F( \alpha(a)E_{\alpha})= \prod\limits_{\alpha}|\alpha(a)| \mu_F(E_{\alpha})= |\rho(a)| \mu(E)$$
where $\rho = \sum\limits_{\alpha \in \Phi(T,N)} \alpha$.
This principle definitely has a generalization to nonsplit groups, but I have no idea how to get started with it.
