Answer (D): Notice that $2024_b = 2b^3 + 2b + 4 = 2(b + 1)(b^2 - b + 2)$, and consider the residue classes of this number modulo 8. If $b \equiv 7 \pmod 8$, then $b + 1 \equiv 0 \pmod 8$, and if $b \equiv 3 \pmod 8$, then $b^2 - b + 2 \equiv 0 \pmod 8$. In each case $2024_b$ is divisible by 16.
In all other cases $2024_b$ is not divisible by 16. Indeed, if $b \equiv 0, 2, \text{ or } 4 \pmod 8$, then $b + 1$ is odd, and $b^2 - b + 2 \equiv b + 2 \pmod 8$, so $2024_b$ is divisible by no power of 2 greater than $2^3$. If $b \equiv 1 \pmod 5$, then $b + 1$ and $b^2 - b + 2$ are both odd multiples of 2, so $2024_b$ is divisible by 8, but not by 16.
Because $2024 = 253 \cdot 8$, there are $253 \cdot 3 = 759$ positive integers $b \le 2024$ that are congruent to 3, 6, or 7 modulo 8. The number 3 must be excluded from this total, because the problem statement requires $b$ to be at least 5. Thus $K = 759 - 1 = 758$, and the sum of the digits of $K$ is $7 + 5 + 8 = 20$.

Notice that $2024_b = 2b^3 + 2b + 4 = 2(b+1)(b^2 - b + 2)$, and consider the residue classes of this number modulo 8. If $b \equiv 7 \pmod{8}$, then $b+1 \equiv 0 \pmod{8}$, and if $b \equiv 3 \pmod{8}$, then $b^2 - b + 2 \equiv 0 \pmod{8}$. In each case $2024_b$ is divisible by 16.
In all other cases $2024_b$ is not divisible by 16. Indeed, if $b \equiv 0, 2, \text{ or } 4 \pmod{8}$, then $b+1$ is odd, and $b^2 - b + 2 \equiv b + 2 \pmod{8}$, so $2024_b$ is divisible by no power of 2 greater than $2^3$. If $b \equiv 1 \pmod{8}$, then $b+1$ and $b^2 - b + 2$ are both odd multiples of 2, so $2024_b$ is divisible by 8, but not by 16.
Because $2024 = 253 \cdot 8$, there are $253 \cdot 3 = 759$ positive integers $b \le 2024$ that are congruent to 3, 6, or 7 modulo 8. The number 3 must be excluded from this total, because the problem statement requires $b$ to be at least 5. Thus $K = 759 - 1 = 758$, and the sum of the digits of $K$ is $7+5+8 = 20$.