Whenever we are given chip description as $xK\times y$, address lines required are $logxK$ and data lines required are $y$.
In this question, the size of chip is $8K\times 4$, which is a nibble, meaning one memory cell is storing only half of data. We'll need two simultaneous/consecutive cell access to output one byte of data, effectively halving storage capacity of the chip.
This means the memory capacity of chip is actually $\frac{8K}{2}=4K$. For $4K$ memory we need $log4K = 12$ address lines.
There are $4\times6 = 24$ such chips. We need $\left \lceil log 24 \right \rceil=5$ bits to select one out of $24$ chips.
Total address lines are $12 + 5 = 17$. OPTION D.