Method I:

N = $9999 \ldots 99$ times

Any digit repeated $(P - 1)$ times is divisible by $P$, where $P$ is a prime number $> 5$.

So, $9999 \ldots$repeated $13 - 1 = 12$ times will be divisible by $13$.

So, $9999 \ldots$repeated $12 \times 8 = 96$ times will be divisible by $13$.

That is, Remainder $[9999 \ldots 96 \text{ times } / 13] = 0$

Or Remainder $[9999 \ldots (96 \text{ times}) 000 / 13] = 0$

So, we just need to find out the remainder when we divide the remaining three digits by $13$.

Remainder $[999 / 13] = 11$

Hence, option (a) is correct.

Method II:

We can analyze the pattern of remainders. Remainder $[9/13] = 9$ Remainder $[99/13] = 8$ Remainder $[999/13] = 11$ Remainder $[9999/13] = 2$ Remainder $[99999/13] = 3$ Remainder $[999999/13] = 0$ This pattern can be seen getting repeated thereafter too. Remainder $[9999999/13] = 9$ Remainder $[99999999/13] = 8$ ...and so forth.

So, if total number of 9s is six, twelve, eighteen, ......., ninety, ninety six, etc., the remainder is 0. So, if the number has ninety seven 9s, the remainder is 9. [Following the pattern] So, if the number has ninety eight 9s, the remainder is 8. So, if the number has ninety nine 9s, the remainder is 11.

Therefore, the answer is 11.

Hence, option (A) is correct.