Answer     D

Firstly, $2008^2$ must end with $4$. Meanwhile the units digit of $2^k$ repeats every $4$ numbers: $2$, $4$, $8$, $6$, $2$, $\cdots$. This means that the end digit of $2^{2008}$ must end with $6$. Therefore, $k$ ends with $0$ which implies $k^2$ ends with $0$.

To determine the last digit of $2^k$, it is sufficient to compute $k\pmod{4}$ because we can utilize the repeating pattern observed above. It is easy to see that $k\equiv 0\pmod{4}$. Therefore, $2^k$ will end with $6$.

Hence $k^2 + 2^k$ ends with $\boxed{6}$.