 Solution

Color every edge independently and uniformly with one of $k$ colors. For each unoriented simple four-edge path $P$, let $A_P$ be its repetition event. Its four distinct edges must satisfy two prescribed color equalities, so $\mathbb{P}(A_P)=k^{-2}$.

Events with edge-disjoint paths depend on disjoint random variables. A fixed edge lies on at most $4{\Delta }^3$ unoriented four-edge paths: count its four possible positions and two orientations, bound the remaining choices by ${\Delta }^3$, then divide the oriented path count by two. Thus the union of all paths sharing any of $P$'s four edges has size at most $16{\Delta }^3$. This union includes $P$ itself, so a dependency graph has $d+1\le 16{\Delta }^3$.

For $k\ge 8{\Delta }^{3/2}$,The symmetric Lovász local lemma yields positive probability of avoiding every event. Therefore such an edge coloring exists. If there are no four-edge paths the conclusion is immediate.