Quick Sort Equal to or Less Than
It doesn't really matter which way you decide to break the tightness. You could put all the elements less than or equal to the pivot in the first list and the ones larger than the pivot in the second list, or you could put only the elements smaller than the pivot in the fist list and the ones larger than or equal to the pivot in the second list.

You could even arbitrarily pick the list that will contain an element equal to the pivot (except the pivot itself), on an element-to-element basis. No matter how you will assign these elements to the lists, the algorithm will still work.

Another solution that slightly breaks the algorithm's description you gave is to put all elements smaller than the pivot in the first list $L_1$, all those larger than the pivot in the second list $L_2$, and insert as many copies of the pivot $x$ as there are elements equal to $x$ (including the pivot itself) between the sorted versions of $L_1$ and $L_2$.

As a final note: the pseudocode reported in your textbook seems wrong. You are supposed to recursively repeat the whole procedure on $L_1$ and $L_2$, not just steps $1$ to $4$ (otherwise you will never reach a recursion depth larger than $1$).