A subgroup $H$ of $G$ is said to pronormal if for all $g\in G$, there exists $x \in \langle H, H^a \rangle$ such that $H^x = H^g$
Now if $H$ and $K$ are pronormal in $G$, it does not follow that $\langle H, K\rangle$ is pronormal in $G$. The counterexample is: Consider the primitive group $G = NK$ where $N=\text{Soc}(G)$ has order 49 and $K \cong S_3$. Let $P$ be a Sylow $3$-subgroup. Then there exists a subgroup $H$ of order 21 such that $H = (H \cap N)P$.
let $ 1_G \neq n \in H \cap N$. Then as any Sylow subgroups of a group is pronormal, $P$ and $P^n$ are pronormal in $G$. I claim $H = \langle P, P^n \rangle$.
It is clear that $P \leq H$ and $P^n \leq H$ so that $ \langle P, P^n \rangle \leq H$. Since the Sylow $3$-subgroups have order 3, $P \cap P^n = \{1_G\}$. thus $PP^n$ has 9 elements and $PP^n \subseteq \langle P, P^n \rangle$. By Langrange, this forces $|\langle P, P^n \rangle| = 21 = |H|$. Hence the claim is proved, and therefore $H$ is the join of two pronormal subgroups of $G$.
Question: I need to show that $H$ is not pronormal in $G$. i.e to find some $g\in G$ such that $H$ and $H^g$ is not conjugate in $\langle H, H^g \rangle$.