Solution:

The answer is (B). Let us write the three equations in the form $(a+c)(a-c)=1024$, $(b+d)(b-d)=1024$ and $(a+c)(b+d)=1024$. Note that $a+c$ and $a-c$ cannot be zero, since $(a+c)(a-c)=1024$. Comparing the first and third equations we then get $b+d=\frac{1024}{a+c}=a-c$, which substituted into the second gives $(b-d)(a-c)=1024=(a+c)(a-c)$, from which $b-d=a+c$; subtracting (term by term) this equation from $b+d=a-c$ we find $2d=-2c$. The numbers $c$ and $d$ are therefore either both equal to zero, or of opposite sign; since by hypothesis $c$ and $d$ are both non-negative, we must then have $c=d=0$. The initial system therefore reduces to the equations $a^{2}=b^{2}=a b=1024=32^{2}$, $c=d=0$, which admit as the unique solution in non-negative integers the quadruple $(a, b, c, d)=(32,32,0,0)$.