Rank nullity theorem -bijection

$(d)$ \begin{align} R(T)+N(T)=n\tag1\\ R(T)-N(T)=n\tag2 \end{align} We know $(1)$ holds anyway due to rank-nullity theorem. Thus solving $(1)$ and $(2)$ we get $R(T)=n$ and $N(T)=0$ which implies $T$ is a bijection.

$(c)$ is an incorrect choice. Consider $T:\mathbb{R}^2\to \mathbb{R}^2$ such that $T(x_1,x_2)=(0,x_1).$ $R(T)+N(T)=n$ will anyway hold. But this $T$ is not a bijection.

In addition to option (d), option (b) is correct. If $n = 0$ then the consequence, $T$ is bijective, is true. On the other hand if $n \ne 0$ then the premise, that the rank equals the nullity, is false. Either way the implication is true.

The question isn't asking you which of the options will be the case if $T$ is a bijection, it's asking which of those prove that $T$ is a bijection. Your reasoning for the first two being incorrect is valid. $c$ certainly doesn't imply that $T$ is a bijection, since it is just the statement of the rank-nullity theorem (for maps from a finite dimensional vector space to itself), which holds for non-invertible transformations. $d$ does imply that $T$ is a bijection, since if the nullity were greater than zero then the rank would be greater than $n$, which is impossible; thus, $T$ is injective, and therefore surjective.