Let 
be a Cayley diagram of group $G$. Let $H$ be the orbit of element p. Is $H$ a normal subgroup of $G$?
Is there a simple way to check that because going by definition seems complicated. I tried to see if the left coset and right coset are equal, but don't really know how to see it from the graph. I know that orbit of p is $\{e, p, a, l\} $. Is the right coset $fH = \{ f, k, o, b\}. $How do I find the right cosets? Should I look at each element of $G$ as a combination of f-s and m-s?
First, remember that from the Cayley graph we can read off the generators of the group by taking one step in each different colored edge starting from the identity $e$. So from your graph, we see that the group is generated by $m$ (red) and $f$ (blue). Further, we find that $p = mf$ since we can get to $p$ from $e$ by taking a red edge then a blue edge. I believe by the orbit of p you mean the subgroup generated by $p$, that is, all elements of the form $p^k$ for $k\in\mathbb{Z}^+$, if this is indeed what you mean, then you've found the right $H$. Though when you talk about orbits you need to specify what action you're referring to. Assuming you have the right $H$, then it suffices to look at $ghg^{-1}$ for various $g\in G$ and $h\in H$ and check if $ghg^{-1}\in H$. Consider $fpf^{-1}$, we can get this from the graph by starting from $e$ and taking the path blue red blue (blue$)^{-1}$ which leaves us at $n\notin H$. Thus H is not normal.