If $E$ has split multiplicative reduction, then $E$ has a $p$-adic uniformization by a Tate curve, and so the $p$-torsion is given as a $G_{\mathbf{Q}_p}$-module by $\{q^{1/p},\zeta_p\}$. In particular, for $p > 2$, there is a $\mathbf{Q}_p$-rational $p$-torsion point exactly when $q$ is a perfect $p$th power. (When $p = 2$, you can also explicitly work out the answer from this description.)
If $E$ has non-split multiplicative reduction, then (still for $p > 2$) you never have $\mathbf{Q}_p$-rational $p$-torsion, because the corresponding Galois representation is the one above coming from $\{q^{1/p},\zeta_p\}$ but then twisted by a quadratic unramified character.
Note that since
$$j = \frac{1}{q} + 744 + 196884 q + \ldots,$$
we have $v(1/j) > 0$ and
$$q = \frac{1}{j} + \frac{744}{j^2} + \ldots,$$
For $q$ to be a $p$-th power we must have $v(q) \equiv 0 \pmod p$ and thus $v(j) \equiv 0 \pmod p$ as well. Hence $q$ is a $p$th power under these conditions precisely when $j$ is a $p$th power. Thus one has:
Let $E/\mathbf{Q}_p$ have multiplicative reduction, and suppose that $p > 2$.
- If $E/\mathbf{Q}_p$ has non-split multiplicative reduction, then there are no rational $p$-torsion points.
If $E/\mathbf{Q}_p$ has non-split multiplicative reduction, then there are no rational $p$-torsion points.
If $E/\mathbf{Q}_p$ has split multiplicative reduction, then there are $p$-torsion points if and only if the parameter $q \in \mathbf{Q}^{\times}_p$ is a perfect $p$th power, which is equivalent to $j \in \mathbf{Q}^{\times}_p$ being a $p$th power.
- If $E/\mathbf{Q}_p$ has split multiplicative reduction, then there are $p$-torsion points if and only if the parameter $q \in \mathbf{Q}^{\times}_p$ is a perfect $p$th power, which is equivalent to $j \in \mathbf{Q}^{\times}_p$ being a $p$th power.
