So we have here the remainder of 2^p divided by 10. That basically means figuring out the last digit of the number 2^p.
1. If s is even, that means that p is even, so p is smth like p = 2x. This means that 2^p = 2^(2x) = (2^2)^x = 4^x.
Now we need to remember that 4 raised to any power has just two possible last digits, and those are 4 and 6. Let me give you a clearer explanation:
4^1 = 4 - last digit 4
4^2 = 16 - last digit 6
4^3 = 64 - last digit 4 (since 6*4 = 24, with last digit 4) and so on.
We can generalize this like so:
4^(even number) - last digit 4
4^(odd number) - last digit 6.

So the last digit of 4^x is either 4 or 6, but since we do not know if x is even as well, then we cannot tell.

2. is a bit more useful, basically telling us that p is divisible by 4, with p = 4t. That means that 2^p = 2^(4t) = (2^4)^t = 16^t. Now here's the interesting part: Raise any number ending in 6 to any possible power and the resulting number will always have the last digit 6. This is because 6*6 = 36, which also ends in 6. This is also true for numbers ending in 5 (5*5 = 25, ending again in 5) and numbers ending in 1 (1*1 = 1, same way of thinking).

So this means that 2^p = 16^t will always have 6 as the last digit, giving us the remainder 6 when divided by 10. So answer will be B.