Let's pursue Jochen's idea. We assume $A \ne \emptyset.$
Let
$$ \varphi(t) = \begin{cases}
e^{-\frac{1}{t}} &\text{if $ t>0$}\\
0 &\text{otherwise.}
\end{cases}$$
This function is $\mathcal C^{\infty}$, and $0 < \varphi(t)$ iff $0<t.$
Define $\rho$ as
$$\rho(x) := k \varphi(1- \|x\|_2^2)$$
where $k$ is chosen so that $\int \rho(x)\space dx =1.$
$\rho$ is $\mathcal C^{\infty}$, $0 \le \rho$, and $0< \rho(x)$ iff $\|x\|_2^2 < 1.$
Let $\delta > 0$. Set $\rho_{\delta}(x) = \frac{1}{\delta^n} \rho(\frac{x}{\delta})$. Note that $0<\rho_{\delta}(x)$ iff $ \|x\|_p < \delta$, and $\int \rho_\delta = 1.$
Let $\mathscr O = A+B_2(0,\delta)$, and consider the function
$$ f = I_{\mathscr O} \ast \rho_{\delta} .$$
Clearly $0 \le f \le 1$ and $f$ is $\mathcal C^{\infty}.$
If $a \in A$, then $B_2(a,\delta) \subset A+B_2(0,\delta) = \mathscr O $, so
$ I_{\mathscr O } \ast \rho_{\delta}(a) = 1.$
If $x \notin A+B_2(0,\delta)$ then
$B_2(x,\delta) \cap (A + B_2(x,\delta)) = emptyset$,
so
$ I_{\mathscr O } \ast \rho_{\delta}(x) = 0.$
For derivatives of any order:
$$ D^\nu f(x) =
D^\nu \int_{\mathscr O} \frac{1}{\delta^n} \rho(\frac{x-y}{\delta}) dy = \frac{1}{\delta ^{|\nu|}} \int_{\mathscr O} \frac{1}{\delta^n} (D^\nu\rho)(\frac{x-y}{\delta}) dy
$$
so
$$ |D^\nu f(x)| \le
( \int |(D^\nu\rho)(y)| dy)\frac{1}{\delta ^{|\nu|}}.
$$
Finally, for any $p$, there is a constant C (depending on $p$ and $n$) such that $B_2(0,1) \subset B_p(0,C_p)$. Therefore:
$$ A \subset A+B_2(0,\delta) \subset A+B_2(0,2\delta) \subset A+B_p(0,2C_p\delta) = A + B_p(0,\epsilon) = A^{\epsilon} $$
if we take $\delta = \epsilon /(2C_p) .$
With this choice of $\delta$, the function $f$ satisfies all the requirements.
$$\dots\dots\dots$$
In order to apply the result without distinguishing the dimention $n$ of the underlying space, it is of interest to consider the following lemma.
$\mathbf{Lemma}$: Given $ p \in \{1,2\} \cup (3,+\infty]$, there is $\rho:\mathbb R^n \rightarrow \mathbb R$ such that:
i) $\rho$ is $\mathscr C^3$
ii) $0 \le \rho$
iii) $\rho(x) > 0$ iff $\|x\|_p < 1$
(iv) and, by re-scaling, $\int \rho(x) \space dx = 1$)
With this choice of $\rho$ the original argument can be carried without resorting to the 2-norm.
$\mathbf{Proof}$: With $\varphi$ as above, if $p \in \{2\} \cup (3,+\infty)$, define $\rho$ as
$$\rho(x) := \varphi(1- \|x\|_p^p) .$$
With this choice, $\rho$ satisfies i,ii,iii. The condition $p>3$ is needed to justify the claimed smoothness of $\rho.$
If $p=1$ or $p=\infty$, note that the open unit ball in $(\mathbb R^n , \|.\|_p)$ is a polytope of the form $P=\{x| v_i^{'}x<1\}.$ When $p=\infty$, $\{v_i\} = \{\pm e_j\}$ where $\{e_j\}$ is the canonical (vector space) bases of $\mathbb R^n$. When $p=1$, $\{v_i\} = \{-1,1\}^n$. In these cases, take $\rho$ as
$$ \rho(x) = \prod_i \space \varphi(1-v_i^{'}x) . \blacksquare$$
Yes, such a $\kappa$ exists for every compact set $K$ with non-empty interior. Here is an abstract linear-algebra argument.
Let $V$ be the real vector space spanned by the multi-indices $\alpha$ with length at most $m$.
We have a linear map $A\colon V \to \mathbb R^K$, sending $(c_\alpha)$ to the function $x\to \sum_\alpha c_\alpha x^\alpha$. Since different polynomials cannot agree on a set with non-empty interior, this map is injective. There is then a finite subset $F\subset K$ such that the composition $V\xrightarrow A\mathbb R^K\to\mathbb R^F$ is a linear isomorphism (for example, pick the elements of $F$ one by one, making sure that the rank of the composition increases by 1 each time).
Now $\kappa$ exists by the fact that all linear isomorphisms between finite-dimensional vector spaces are bicontinuous.
If you want an explicit value for $\kappa$, you can use discrete derivatives (see https://en.wikipedia.org/wiki/Finite_difference) to write $c_\alpha$ as a linear combination of some values of $P$.
Best Answer
The answer is no. E.g., let $d=1$, $K=[0,1]$, and, for $x\in K$, $$P(x):=T_n(x):=n\sum_{0\le k\le n/2}\frac{(-1)^k}{n-k}\binom{n-k}k2^{n-2k-1}x^{n-2k} =\cos(n\arccos x),$$ the $n$th Chebyshev polynomial.
Then $\|P\|_{\infty;K}\le1$, whereas (say) for $k=0$ we have $\|P_k\|_{\infty;K}=2^{n-1}\to\infty$ as $n\to\infty$.