We know that 20 multiplied by an odd or even number will be even (ex 20*7 = 140; 20*4 = 80), so p can be odd or even.
Since 20*anything is always even, we need 3q to be odd in order to sum to an odd number, therefore q must be an odd number

Since p can be even or odd, we'll take two instances: p=2 and p=3
Since q must be odd, we'll say q=1

we then will look at all instances where the answer can be even, and eliminate those options
Example:
A) p - q
p=3, q=1
3-1=2, which is even, so we cross off option A

B) p + 2q
p=3, q=1
3+2(1) = 4, also even

Our answer, option D) 2p + q^2
Instance 1 where p=2: 2(2)+1^2 = 5 odd
Instance 2 where p=3: 3(2)+1^2 = 7 odd

Since D is odd regardless of weather p is even or odd, it is our final answer

Ah, I see. I was being careless. Thanks so much for the clarification!