No. Consider the graph $G = (V, E)$ with $$V = \{0,1,2,3,4,5,6\}\quad \text{and} \quad E = \{\{0,1\}, \{1, 2\}, \{0,3\}, \{3, 4\}, \{0,5\}, \{5, 6\}\}.$$ Note that its line-graph has three vertices of degree 1. Therefore $L(G)$ has no Hamiltonian path.
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