They are equal up to sign. 

If $F\to E\to B$ is a Hurewicz fibration, where $B$ is well-pointed, then we have a factorization $E\to E/F \to B$ and we have the Barratt-Puppe extension $E/F \to \Sigma F$. This gives a diagram
$$
B \quad \overset{a}\leftarrow \quad E/F \quad \overset{b} \to \quad \Sigma F
$$
and when the transgression is defined it is given by the homomorphism these maps induce on homology.

More precisely,  if  $x  \in H_k(\Sigma F)$ lifts to an element $y$ of $H_k(E/F)$ via 
$b_\ast$, then one defines the transgression of $x$ as $a_*y$.  To make this well-defined, you need to take into account the indeterminacy of the lifts $y$ (this is the image
of the map $H_k(E) \to H_k(E/F)$ and you must quotient out by this indeterminacy--but this won't matter in your case--cf. below).

(You can see this e.g., in McCleary's book, p. 185)


Now consider the path-loop fibration $\Omega X \to PX \to X$. In this case $E/F$ is identified with $\Sigma \Omega X$ and the map $E/F\to \Sigma F$ is identified with
$\pm$ identity. Thus the lift is unique in this case (no indeterminacy) and the transgression is the map $\Sigma \Omega X \to X$ up to sign on homology.