Let $L$ be a Lie algebra and let $\Phi$ be a root system and $\Delta$ be a basis. Let $\Gamma\subset \Delta$. Define, $$P:=H\oplus\displaystyle\sum_{\alpha\in\Phi_+} L_\alpha \oplus\displaystyle\sum_{\alpha\in\Gamma} L_{-\alpha} $$ How to show that $N_L(P)=P$, and how to find $Rad(P)$?
How to prove that $P/Rad(P)$ is isomorphic to the subalgebra generated by all $L_\alpha$ and $L_{-\alpha}$ for $\alpha\in\Gamma$?
Your definition of $P$ is not correct, you should take a subset $\Gamma\subset\Phi$ which is closed under addition. This is then uniquely determined by $\Sigma:=\Gamma\cap\Delta$. The normalizer of $P$ has to be a direct sum of root spaces. But since $H\subset P$ and $[H,X]=\mathfrak g_\alpha$ for each non-zero $X\in\mathfrak g_\alpha$, you see that a root space contained in the normalizer must be contained in $P$.
Next, you can decompose $H=H'\oplus H''$ where $H'$ is joint kernel of the roots in $\Sigma$, while $H''$ is spanned by the brackets $[\mathfrak g_\alpha,\mathfrak g_{-\alpha}]$ for $\alpha\in\Sigma$. Then the sum $R$ of $H'$ with the root spaces $\mathfrak g_\alpha$ with positive $\alpha$ such that $\mathfrak g_\alpha$ is not contained in $P$ is a solvable ideal. On the other hand, the sum of $H''$ and the remaining root spaces is easily seen to be semisimple. This implies that $R$ is the radical of $P$ as well at the statement on $P/R$.
