Answer

The only even prime number is 2, and others are odd (otherwise they would be divisible by 2 and therefore not prime). So if the 4 numbers we draw out of the first 16 primes don't include 2, they must all be odd. Then, even number of odd numbers always sum up to an even number, so our sum is even if 2 wasn't picked. If 2 was picked, then we have one even number and 3 odd numbers. Odd number of odd numbers sums to an odd number, so the three odd primes sum to an odd number. Adding 2 to this still leaves an odd sum. So if 2 was picked, we get an odd sum.

So what's the chance we didn't pick 2? We can find this by counting the number of possible picks excluding 2 and dividing this with the total number of possible picks. If we exclude 2, we are picking 4 numbers out of 15 in $\binom{15}{4} = \frac{15!}{4! 11!}$ possible ways. In total, we can pick 4 numbers out of 16 in $\binom{16}{4}=\frac{16!}{4! 12!}$ ways. Therefore the probability of an even sum is $$\frac{15!\cdot 4! \cdot 12!}{16! \cdot 4! \cdot 11!}=\frac{3}{4}$$.