One has the finite flat group scheme $\mathbb Z/p$ over $\mathcal O_{K_{\mathfrak p}}$
(I write $K_{\mathfrak p}$ for the $\mathfrak p$-adic completion of $K$, and
$\mathcal O_{K_{\mathfrak p}}$ for its integer ring),
as well as
the finite flat group scheme $A[p]$. Giving a $p$-torsion point over $K_{\mathfrak p}$ (and hence in particular over $K$) is the same as giving a closed embedding
on generic fibres:
$(\mathbb Z/p)\_{/ K_{\mathfrak p}} \hookrightarrow A[p]\_{/K_{\mathfrak p}}.$

Raynaud's results imply that this extends to a closed embedding over $\mathcal O_K$:
$\mathbb Z/p \hookrightarrow A[p],$
which is another way of saying the that the non-zero $p$-torsion point has non-zero
reduction.

Just to see concretely what can happen in the situation when $e \geq p-1$, suppose
that $K = \mathbb Q$ and $p = 2$. Then we could have a map
$(\mathbb Z/2)\_{/\mathbb Q_2} \hookrightarrow A[2]_{/\mathbb Q_2}$
which extends to a closed immersion
$\mu_2 \hookrightarrow A[2].$
This would correspond to having a 2-torsion point in the kernel of the reduction map.
(Note that $\mu_2$ has a non-trivial point in char. zero, which collapses down to
the identity in char. two.)

[Added in response to unkwown's comment:]

The point is that one can form the scheme-theoretic closure in $A[p]$ of the image
of $(\mathbb Z/p)\_K,$ which is *some* finite flat subgroup scheme over $\mathcal O_K$
which is embedding as a closed subgroup scheme of $A[p]$ (by construction: we formed it as a scheme-theoretic closure). And it
has $(\mathbb Z/p)\_K$ as its generic fibre (again by construction).

Now when $e < p-1$, Raynaud's results show that this finite flat group scheme has no
choice but to be $\mathbb Z/p$, and so we get a copy of $\mathbb Z/p$ embedding into $A[p]$,
extending the original embedding of generic fibres. Thus the order $p$ point in $A[p](K)$
reduces mod $p$ to an order $p$ point.

But if e.g. $p = 2$, then this scheme-theoretic closure could be $\mu_2$. Now the non-trivial point ($-1$) of $\mu_2(K)$ specializes to the trivial point in char. 2,
and so when we have a copy of $\mu_2$ inside $A[2]$, the non-trivial point of $\mu_2(K)$
lies in the kernel of the reduction mod 2 map.