Possible $p$-torsion subgroup of $E(\mathbb{Q}_p)$, and if there is a theorem to say which case happens when? What is the possible $p$-torsion subgroup of $E(\mathbb{Q}_p)$ for an elliptic curve $E$ over $\mathbb{Q}_p$, and if there is a theorem to say which case happens when?
 A: If $p$ is odd then $E(\mathbb{Q}_p)[p]$ is either trivial or one copy of $\mathbb{Z}/p\mathbb{Z}$ and that will be all of the $p$-primary torsion. For $p=2$, all of $E[2]$ can be defined over $\mathbb{Q}$ and hence $\mathbb{Q}_2$.
For odd primes: If the reduction is good and the reduction has no $p$-torsion (not "anomalous" as it is called) then there are no $p$-torsion over $\mathbb{Q}_p$. For good anomalous reduction both cases can occur (see below). If the reduction is multiplicative there are no $p$-torsion points, while for additive reduction it is more complicated again, especially if $p=3$. This question explains how to distinguish algorithmically between the two cases for odd primes. 
A simple argument why one cannot have all $p$-torsion rational over $\mathbb{Q}_p$ in the odd case is that the $p$-th roots of unity $\mu_p$ are not in $\mathbb{Q}_p$ and so the Weil pairing provides a reason. 
I recommend section VII.3 in Silverman 1 or chapter 11 in Cassels' "Lecutre on Elliptic Curves".
To explain why we don't expect a theorem that will distinguish easily the two cases look at $E_1: y^2 + y = x^3 - x^2$ and $E_2 : y^2 = x^3+3x+3$ both for $p=5$.
They have good reduction with precisely $5$-points in $\tilde{E}(\mathbb{F}_5)$.
In the first case the point $(0,0)$ is a $5$-torsion point, while in the second there are no torsion points in $E_2(\mathbb{Q}_5)$. The sequence
$0\to \hat{E}(p\mathbb{Z}_p)\to E(\mathbb{Q}_p)\to \tilde{E}(\mathbb{F}_p)\to 0$ has a copy of $\mathbb{Z}_p$ on the left and a cyclic group of order $p$ on the right. For $E_1$ the sequence splits for $E_2$ it does not.
A: I was interested in this question myself a while back, particularly for the additive reduction case. I wrote up a little note about my results here. The main result was a nice looking numerical criterion (in terms of a Weierstrass equation):

Let $E/\mathbb{Q}_p$ be an elliptic curve with additive reduction, and assume that it is 
given by a minimal Weierstrass equation over $\mathbb{Z}_p$:
$$y^2 + a_1 xy + a_3 y = x^3 + a_2 x^2 + a_4 x + a_6,
$$
where the $a_i$ are contained in $p \mathbb{Z}_p$ for each $i$. Then the group $E_0(\mathbb{Q}_p)$ (points of good reduction) is topologically isomorphic to $\mathbb{Z}_p$, except in the following four cases:

*

*$p = 2$ and $a_1 + a_3 \equiv 2 \pmod{4}$;

*$p = 3$ and $a_2 \equiv 6 \pmod{9}$;

*$p = 5$ and $a_4 \equiv 10 \pmod{25}$;

*$p = 7$ and $a_6 \equiv 14 \pmod{49}$.

In each of the above four special cases, $E_0(\mathbb{Q}_p)$ is topologically isomorphic to $p \mathbb{Z}_p \times \mathbb{Z}/p \mathbb{Z}$, where the second factor in the direct product has the discrete topology.

This completely answers your question for $p>3$, and for $p=3$ if the points of good reduction contain a point of order $p$ (bearing in mind the remark of Chris Wuthrich that there cannot be full $p$-torsion for $p>2$).
However, both for $p=2$ and $p=3$ we could still expect a contribution to the $p$-torsion coming from the points of bad reduction. In order to control this, one needs to know the Kodaira type of the special fibre of the Néron model. If the rational points on the special fibre contain $p$-torsion (which can be decided with Tate's algorithm), then as in Chris Wuthrich's answer there is again the question whether or not this lifts to the $p$-adic points. One can possibly still say some smart things about this case, but I have forgotten most of my thoughts about this subject unfortunately. (But the most sensible thing to do for this special case seems to me to look at the $2$- and $3$-division polynomials.)
A: 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 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.

