Let the prime be $p$. Then we have $p = a+b$ and $p=c-d$, where $a$, $b$, $c$, and $d$ are also primes and WLOG $a>b$. It is easy to see that $p=2$ renders no solutions, so we must have $p$ be odd. For $a$ and $b$ to add to an odd number, one must be even and the other odd, so $b = 2$ (since 2 is the smallest prime). Similarly, for $c$ and $d$ to subtract to an odd number, one must be even and the other odd, so $c=2$ (since 2 is the smallest prime). So we now have $p = a+2 = d-2$. This condition is equivalent having three odd consecutive primes, $a$, $p$, and $d$ (in order). Since (pairwise), they differ by 2 and 4, none of which are multiples of 3, one of $a$, $p$, and $d$ must be a multiple of three. (this can also be shown formally with a mod argument). Since they are primes, this is only true if the three numbers are 3, 5, and 7. So the only prime is $\boxed{5}$. $\blacksquare$

Let the prime be $p$. Then we have $p = a+b$ and $p=c-d$, where $a$, $b$, $c$, and $d$ are also primes and WLOG $a>b$. It is easy to see that $p=2$ renders no solutions, so we must have $p$ be odd. For $a$ and $b$ to add to an odd number, one must be even and the other odd, so $b = 2$ (since 2 is the smallest prime). Similarly, for $c$ and $d$ to subtract to an odd number, one must be even and the other odd, so $c=2$ (since 2 is the smallest prime). So we mow have $p = a+2 = d-2$. This condition is equivalent having three odd consecutive primes, $a$, $p$, and $d$ (in order). Since (pairwise), they differ by 2 and 4, neither of which are multiples of 3, one of $a$, $p$, and $d$ must be a multiple of three. (this can also be shown formally with a mod argument). Since they are primes, this is only true if the three numbers are 3, 5, and 7. So the only prime is $\boxed{5}$. $\blacksquare$

Let us suppose that $p=k+l=a-b$, then we can easily see that $p$ must be odd. Now, for the generation of other primes, one of the primes in both equation must be $2$. Now, we get three consecutive odd numbers as prime but one of the members must be divisible by $3$, so we can only have $3,5,7$ as consecutive odd primes. Hence the only prime that satisfies the condition is $\boxed{p=5}.\blacksquare$