So $D_4$ is the symmetries of the square, I am told $L=<P>$ which is really just $L=${$P,I$} where $I$ is the identity and $P$ is whatever flipping of the square you desire. Now, in my text it says $L$ is normal in $D_4$ if $glg^{-1}$ is in $L$ for all $g\in D_4, l\in L$.
Clearly I can pick $g=R$ so that $g^{-1}=R^3$, and $l=P$ I get $RPR^3=VR^3=O$ and $O \notin L$
So it cant be $L$ is normal in $D_4$
Now I'm asked to construct $D_4/L$ but I don't understand how. I do know you would use the cosets and I have calculated the right and left cosets which do not match so that further tells me $L$ is not normal. Is it possible to do the table?
You're right. It's not normal.
To make sense of $G/H$ for some $H<G$, one needs to know whether one is considering left cosets or right. Normality ensures that $gH=Hg$ for all $g\in G$; however, without this normality, it is still possible to construct a Cayley table, provided the knowledge of which cosets to use.