If $L$ is a normal subgroup of $G$ and if $L=HK$ (where $H$ is normal in $G$ and $K$ is any subgroup), then must $K$ also be normal? I think the answer is yes, but I just want to ask in case I'm missing something...

We know that $aKH=KHa$ for all a in G. But we also know that $aH = Ha$, so $KHa = KaH$. 

So $aKH = KaH \implies aK=Ka$. 

Thanks in advance
This is false. Take G=L=H to be nonabelian of order 6, and K of order 2.

Notice that aKH = aKG = G, and that KaH = KaG = G, so aKH = KaH only tells you G = G, nothing about K or a. You cannot generally cancel subsets.
So $aKH=KaH  \ then  K^{-1}a^{-1}Ka\in H$\ 
It seems this way cannot be Ok. you can use this way :
$K=H^{-1}L$ is normal subgroup ,because $H\  L$ are both normal .