Convergence in probability and uniform integrability implies convergence in Lp

You should use the Vitali convergence theorem (see eg Wikipedia https://en.wikipedia.org/wiki/Vitali_convergence_theorem ):

If $\lim_{a\to\infty} \int_{\{ x \mid\  |X_n(x)|>a}|X_n|^p \to0$ uniformly in $n$ then whenever $A_m$ is a sequence of sets with $P(A_m)\to 0$ you have that $\lim_{m\to\infty}\int_{A_m} |X_n|^p\to0$ uniformly in $n$.

Apply this like follows. First consider:

$$\int_{\{|X_n-X|>1\}} |X_n-X|^p ≤ \left((\int_{\{|X_n-X|>1\}} |X_n|^p)^{1/p}+(\int_{\{|X_n-X|>1\}}|X|^p)^{1/p}\right)^p$$
Since $X_n\to X$ in probability oyu have that $P( |X_n-X|>1)\to0$ as $n\to\infty$, whence by Vitali the first summand on the right goes to $0$. The second summand goes to $0$ since $X\in L^p$. So you find some $N_1$ after which this sum is $<\epsilon$.

Now look at:
$$\int_{\{|X_n-X|≤1\}}|X_n-X|^p ≤ \sum_{m=0}^\infty 2^{-p\,m} P(2^{-m}≥|X_n-X|>2^{-m})$$
This is far more standard. Note that $P( .... )≤1$ always and that each summand goes to $0$ with $n\to\infty$. So choose some $M$ for which
$$\sum_{m=M}^\infty  2^{-pm}<\epsilon$$
and then some $N_2$ so that for $n>N_2$ you have
$$P(|X_n-X|>2^{-M})<\epsilon$$
then:
$$\sum_{m=0}^\infty 2^{-pm}P(2^{-m}≥|X_n-X|>2^{-m}) = \sum_{m=0}^{M-1} ... + \sum_{m=M}^\infty ... ≤ \epsilon + \epsilon$$

Putting it together:

$$(\|X_n-X\|_p)^p= \int_{\{|X_n-X|≤1\}}|X_n-X|^p + \int_{\{|X_n-X|>1\}}|X_n-X|^p≤ \epsilon + \epsilon + \epsilon$$

whenever $n>\max(N_1,N_2)$ which yields the result since $\epsilon$ is, as always, arbitrary.