Solution:

Note first that $$\dfrac{ S_{2023} - S_{2019}}{2^{2019}} = x_{2019} +$$$$2x_{2020} + 4x_{2021} + 8x_{2022}.$$ Therefore, we should attempt to find $S_{2019}, S_{2023}.$ Also, note that $$0 \leq S_n \leq \sum_{k=0}^{n-1} 2^k < 2^n.$$

Now, since $7S_n \equiv 1 \pmod{2^n},$ we know $$7S_n = m2^n +1 \implies$$$$S_n = \dfrac {m2^n+1}{7}$$ for some integer $m.$ Also, since $0 \leq S_n < 2^n,$ we know $$0 \leq \dfrac {m2^n+1}7 < 2^n.$$ This means $0 \leq m < 7.$ Now, we find $m$ such that $m2^n + 1$ is divisible by $7.$ This makes $m2^n + 1 \equiv 0 \mod 7.$

If $n=2019,$ then $$0 \equiv m2^{2019} + 1 \equiv$$ $$m(2^3)^{673}+1 \equiv$$ $$m8^{673} + 1\equiv m+1,$$ so $m=6.$ This makes $$S_{2019} = \dfrac {6(2^{2019})+1}{7}.$$

If $n=2023,$ then $$0 \equiv m2^{2023} + 1$$ $$\equiv m(2^3)^{674}\cdot 2+1$$ $$\equiv m8^{674}\cdot 2 + 1$$$$\equiv 2m+1,$$ so $m=3.$ This makes $$S_{2023} = \dfrac {3(2^{2023})+1}{7}.$$

Our answer is $$\frac{S_{2023}-S_{2019}}{2^{2019}}=\frac{1}{2^{2019}}\left(\frac{3\cdot2^{2023}+1}{7}-\frac{6\cdot2^{2019}+1}{7}\right)=\frac{42\cdot2^{2019}}{7\cdot2^{2019}}=6.$$

Thus, the correct answer is A .