For conditions: $a>0, b>0, a>b$ & $a^2$ − $b^2$ to be prime number
=> a & b must be consecutive integer i.e., $(a-b)$ = $1$ (for explanation see edit)
thus, $a^2$ − $b^2$ = $(a-b)*(a+b)$ = $(a+b)$.
Hence, option $(B)$ is correct.
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EDIT: factors of prime number: that number itself & 1.
since ($a^2$-$b^2$) have two factors $(a+b)$ & $(a−b)$ and if $(a-b)$ will be any number other than 1 than with given conditions $[a>b,a>0,b>0]$ , ($a^2$-$b^2$) will never be prime number.
so only choice is $(a−b)$ = 1.