Just noticed I misread the problem. The proof I sketch below is if $p_n(x) = \left\{
\begin{array}{l l}
1 & \quad \text{$x=1+1/n$}\\
0 & \quad \text{otherwise}
\end{array} \right.$ but a similar proof holds for the OP's question.
Let $p(x)=0$. Pick any value of $x \notin \{1+1/n : n=1,2,3,\ldots\}$. Then $p_n(x)=0$ so $|p_n(x)-p(x)|=0$. Note that $1 \notin \{1+1/n : n=1,2,3,\ldots\}$ so $p_n(1)=p(1)=0$.
Now pick any $x\in \{1+1/n : n=1,2,3,\ldots\}$. Let $x=1+1/m$, for $n\geq m+1$, $p_n(1+1/m)=0$.
It is even further interesting that although the pmf converges to 0, the corresponding CDF does not!
$F_n(x)=P(X_n\leq x) \overset{D}{\longrightarrow} F(x) = \left\{
\begin{array}{l l}
0 & \quad \text{$x<1$}\\
1 & \quad \text{$x\geq 1$}
\end{array} \right.$ at all points of continuity of $x$. Therefore $X_n \overset{D}{\longrightarrow} X$ where $P(X=1)=1$. Even though the density functions converged...the corresponding probability mass functions do not!