Let $G$ be a connected, reductive group over a $p$-adic field $k$, let $P$ be a parabolic subgroup with Levi $M$ and radical $N$.  Let $K$ be a maximal open compact subgroup of $G$ in good position relative to $P$, so that $G = PK$.  Under what circumstances do we have $P \cap K = (M \cap K)(N \cap K)$?

My motivation is the following: it is a common convention, e.g. in Harish-Chandra's notes, to normalize a left or right Haar measure on each closed subgroup $H$ of $G$ so that $H \cap K$ has measure one.  Supposing we do this for $M$ and $N$, then we get a left Haar measure $d_lp$ on $P$ satisfying:

$$\int\limits_P f(p)d_lp = \int\limits_M \int\limits_N f(mn) dndm$$
My question is then equivalent to the assertion that the Haar measure on $P$ in this way follows the convention that $P \cap K$ has measure one (taking $f = \operatorname{Char} P \cap K$).  

I'm not even sure about the unramified case.  Here $G$ comes from a smooth affine group scheme $\mathcal G$ over $\mathcal O_K$, and $K = \mathcal G(\mathcal O_K) = G(\mathcal O_K)$.  So should $P, M, N$ as $\mathcal P, \mathcal M, \mathcal N$.  We would have $P \cap K = (M \cap K)(N \cap K)$ if the product map $M \times_k N \rightarrow P$ comes from an isomorphism of $\mathcal O_K$-schemes $\mathcal M \times_{\mathcal O_K} \mathcal N \rightarrow \mathcal P$.