Given a path, just iterate along it and find the node with highest capacity.
Distilled from the comments, let us consider this more interesting question:
Given a network of nodes $N$ connected by links $E$ with limited bandwidths $b : E \to \mathbb{N}$, find a path from $s$ to $t$ with maximum bandwith.
In other words, you are looking for an $s$-$t$-path $p = (s, v_1, \dots, v_k, t)$ that maximises
$\qquad \displaystyle B(p) := \min\ \{ b(s,v_1), b(v_k, t) \} \cup \{b(v_i,v_{i+1} \mid i = 1, \dots, k-1\} $
among all $s$-$t$-paths.
It is noteworthy that cycles do not change path bandwith $B$, so we can restrict ourselves to minimal paths in this respect, that is look at $B$-optimal spanning trees.
Note furthermore that among all $B$-optimal paths, there is (at least) one that has the subpath-optimality property we know from the shortest path problem, that is that all partial paths of an optimal paths are optimal (for the nodes they connect). It does not hold for all $B$-optimal paths¹, but that is not a problem: what we need to know is that there is a $B$-optimal path that can be constructed from a concatenation of $B$-optimal paths. Therefore, we can restrict ourselves to considering only such (partial) paths.
With this we have collected enough justification for a greedy strategy similar to Prim's algorithm: starting in $s$, successively choose edges with maximum bandwith among those edges incident to already visited nodes (that do not connect two already visited nodes). This will discover only $B$-optimal paths and therewith a $B$-optimal $s$-$t$-path (if $t$ is reachable from $s$).
This runs in time $O(|N|^2)$.
Consider this graph:

[source]
All $s$-$t$-paths are optimal, but those that go via the lower node have non-optimal subpaths.