Most outstanding prime

Let the primes be $p,q,r,s$ .And let $p+q+r+s=t$ where $t$ is also a prime.

Suppose all the primes $p,q,r,s$ are odd.Then $p+q+r+s=t$ will be even according as parity as $\text{(Odd+Odd)+(Odd+Odd)=Even+Even=Even}$

But that implies that $t$ is an even prime number.Note that as $p,q,r,s$ are all greater than or equal to 3 (since they are odd),their sum must be greater than or equal to 12.This implies that $t$ is an even prime number greater than 12,contradiction.

Hence $t$ is odd and at least one of the prime $p,q,r,s$ is even.Hence their product $pqrs$ will be even as well.

Since all primes are odd except for 2, if we had all of the primes be odd, since odd+odd=even and even+even=even, the sum of these 4 odd primes would be even so we can not have all of the primes be odd, this means that one of the numbers is 2.Since (even)(odd)=even, the product must be even.Note that the numbers 2, 3, 5, 7 and 2, 3, 11, 7 also work so there is more than one pair of numbers satisfying these limitations

If the sum of the four primes is also a prime, then they cannot all be odd, for there sum would be even; so one of the primes must be 2, and the product is even.