In an eight-digit number there are 4 odd and 4 even positions. In the number 11223344, there are 4 odd numbers (1, 1, 3, 3) and 4 even numbers (2, 2, 4, 4).

Now, the number of ways these 4 odd numbers (1, 1, 3, 3) can be arranged in 4 odd positions = $\frac{4!}{(2 \times 2)} = 6$

Similarly, the number of ways these 4 even numbers (2, 2, 4, 4) can be arranged in 4 even positions = $\frac{4!}{(2 \times 2)} = 6$

So, number of such distinct numbers = $6 \times 6 = 36$

Hence, option C is correct.