Solution:

The main idea is the following claim.

Claim: If $a, b, c$ are in arithmetic progression and have the same number of digits in their binary representations, then $r(a), r(b), r(c)$ cannot be in arithmetic progression in that order.

Proof. Consider the least significant digit that differs in $a$ and $b$. Then $c$ will have the same value of that digit as $a$, which will be different from $b$. Since this becomes the most significant digit in $r(a), r(b), r(c)$, then of course $b$ cannot be between $a$ and $c$.

To finish, we just need to show that if there are 8 numbers in arithmetic progression, which we'll write as $a_{1}, a_{1}+d, a_{1}+2d, \ldots, a_{1}+7d$, three of them have the same number of digits. We have a few cases.

- If $a_{1}+3d < 2^{k} \leq a_{1}+4d$, then $a_{1}+4d, a_{1}+5d, a_{1}+6d$ will have the same number of digits.

- If $a_{1}+4d < 2^{k} \leq a_{1}+5d$, then $a_{1}+5d, a_{1}+6d, a_{1}+7d$ will have the same number of digits.

- If neither of these assumptions are true, $a_{1}+3d, a_{1}+4d, a_{1}+5d$ will have the same number of digits.

Having exhausted all cases, we are done.