Is topological sort of an original graph same as post-ordering dfs of its transpose graph
Let $G$ be a graph and $G'$ be it's transposed version. Your property follows from the following two facts:

1) The order in which vertices are visited by a postorder visit on $G'$ is the reverse of a topological order $\sigma'$ for $G'$ (in fact, that's a standard way to compute a topological order).
You can see that this is true by noticing that if $(u,v) \in G'$ then $v$ must be visited before $u$ can be visited, i.e., $v$ follows $u$ in $\sigma'$.

2) If $\sigma'$  is a topological order for $G'$, then the linear order $\sigma$ obtained by reversing $\sigma'$ is a topological order for $G$.
You can see that this is true because, for every edge $(u,v)$ of $G$, $G'$ contains $(v,u)$ and therefore $v$ precedes $u$ in $\sigma'$. This means that $(u,v) \in G \implies u$ precedes $v$ in $\sigma$.