Application of Slutsky theroem
Solution 1:
Let $$Z_n^{1}=\sum_{j=1}^n(X_j-EX_j)/\sigma_n$$ and
$$Z_n^{2}=\sum_{j=1}^n(X_j-EX_j)/n$$
For the if part, you have $Z_n^1 \overset{d}{\rightarrow} Z^1$, where $Z^1 \sim N(0,1)$ and $Z_n^2 \overset{p}{\rightarrow} 0$. Here $Z_n^2$ converges in distribution to the constant 0. By Slutsky's theorem $\frac{Z_n^2}{Z_n^1}$ converges in distribution to $\frac{1}{Z^1} \times 0$ i.e. the constant 0, which is equivalent to convergence in probability to 0.