The given function is $h(a,b)=(2a+1)2^b-1$
$h:N\times N\to N$
(i) injection:-
$h$ will be one-to-one if, $h(a,b)=h(c,d)\implies (a,b)=(c,d)$
$h(a,b)=h(c,d)$
$\Rightarrow (2a+1)2^b-1=(2c+1)2^d-1$
$\Rightarrow \frac{2a+1}{2c+1}=2^{d-b}$
If $d\neq b$ then RHS will always be even, but LHS is (odd/odd) which is odd.
$LHS = RHS \text{ only if }d=b,$
$\Rightarrow \frac{(2a+1)}{(2c+1)}=1$
$\Rightarrow (2a+1)=(2c+1)\Rightarrow a=c$
$\implies (a,b)=(c,d),\; hence\; h$ is one-one.
(ii) Surjection:-
For $h$ to be onto, there should not be any non-negative integer which can't be generated by the function. i.e. all the elements of co-domain should be mapped by some element of domain set.
Now look at the function it is $(2a+1)2^b-1$
If $b=0, (2a+1)2^b-1=2a \implies$ all non-negative even numbers generated.
If $b\neq 0,(2a+1)2^b-1\implies$ all non-negative odd numbers generated.
So all the numbers of co-domain (i.e. $N$) are mapped by some $(a,b)$ pair. Hence h is onto (surjective).