Suppose $G$ is a group with a split $BN$-pair, and if $N_J$ is the subgroup of $N$ such that $N_J/(B\cap N)=W_J$ (the parabolic subgroup in the Weyl group $W=N/(B\cap N)$). Let $P_I$ denote a standard parabolic subgroup.
Why is it that $P_JnP_K\cap N=N_JnN_K$ for all $n\in N$?
Proposition 2.8.1 of Carter's Finite Groups of Lie Type says $$ \begin{aligned} P_JnP_K &= BN_JBnBN_KN\cap N\\ &\subseteq BN_JnBN_KB\cap N\\ &\subseteq BN_JnN_KB\cap N\\ &=N_JnN_K \end{aligned} $$
the two containments following since $N_JBn\subseteq BN_JnB$ and $N_JnBN_K\subseteq BN_JnN_KB$, respectively.
I can't follow the last equality, why does $BN_JnN_KB\cap N=N_JnN_K$? Writing something like $b_1n_1nn_2b_2=n_3$, I can't rearrange it in a way to conclude the element is actually just in $N_JnN_K$.
I am not completely sure about this, but I give it a try.
I think this follows from the Bruhat decomposition, which roughly says that $(B,B)$-orbits in $G$ are parameterized by elements of $N/T$ where $T$ is a (split) maximal tours. Meaning in your situation, you have $b_1n_1nn_2b_2=n_3$ and thus the two $(B,B)$-orbits which arise from $n_1nn_2$ and $n_3$ are equal. By the Bruhat decomposition, you then have $n_1nn_2t=n_3$ for some $t\in T$. This shows that the element you started with $n_3$ lies in $N_JnN_KT=N_JnN_K$ since $T\subseteq N_K$ - as required.