This one is a simple case of combinining two limits into one:
Theorem: If $X_n\overset{d}{\to}X$ and $\mathcal{L}(Y_n) \overset{w.a.}{\longleftrightarrow}\mathcal{L}(X_n)$ then $Y_n\overset{d}{\to}X$.
Proof: Let $f$ be an arbitrary bounded continuous function. Since $X_n\overset{d}{\to}X$ and $\mathcal{L}(Y_n) \overset{w.a.}{\longleftrightarrow}\mathcal{L}(X_n)$ we have the respective limits: $$\begin{aligned} &\lim_{n \to \infty} \mathbb{E}(f(X_n)) - \mathbb{E}(f(X)) = 0, \\[6pt] &\lim_{n \to \infty} \mathbb{E}(f(Y_n)) - \mathbb{E}(f(X_n)) = 0. \\[6pt] \end{aligned}$$ We therefore obtain the limit of interest: $$\begin{aligned} \text{Limit} &\equiv \lim_{n \to \infty} \mathbb{E}(f(Y_n)) - \mathbb{E}(f(X)) \\[6pt] &= \lim_{n \to \infty} \Big[ \mathbb{E}(f(Y_n)) - \mathbb{E}(f(X_n)) + \mathbb{E}(f(X_n)) - \mathbb{E}(f(X)) \Big] \\[6pt] &= \lim_{n \to \infty} \Big[ \mathbb{E}(f(Y_n)) - \mathbb{E}(f(X_n)) \Big] + \lim_{n \to \infty} \Big[ \mathbb{E}(f(X_n)) - \mathbb{E}(f(X)) \Big] \\[6pt] &= 0+0 = 0. \\[6pt] \end{aligned}$$ This establishes that $Y_n\overset{d}{\to}X$, which was to be shown. $\blacksquare$