Any value $\Phi(z)$ determines a unique multiplicative linear map on polynomials, which has a unique continuous (multiplicative linear) extension onto $\mathcal P(K)$ iff $|\Phi(p)|≤\|p\|$ for all polynomials $p$.
As noted in the question, $\delta_x$ for $x\in K$ determines a character in $\mathcal P(K)$, but there are others.
An example:
Let $K$ be the annulus $\overline{B_2(0)-B_1(0)}$. Since for polynomials/holomorphic functions maximal values can never lie on interior points one has that the maximal value of $p$ on $\overline{B_2(0)}\supset K$ must lie on the circle of radius $2$. In particular $\|p\|_K≥|p(0)|=|\delta_0(p)|$ for all $p$ and $\delta_0$ extends to a character on $\mathcal P(K)$, even though $0\notin K$.
This means that $\Phi_{\mathcal P(K)}\neq \Phi_{C(K)}$. More generally one can see the following:
Lemma If $x\in\mathbb C$ such that there exists a differentiable loop $\gamma: S^1\to K$ so that $\gamma$ has non-zero winding number on $x$, then $\delta_x$ extends to a character.
The proof is the same, let $U$ be the open set containing $x$ with boundary $\gamma$. Since $\bar U$ is compact and polynomials must take the maximal value on the boundary one has by virtue of $\gamma \subset K$ that:
$$\|p\|_K≥\|p\|_\gamma=\|p\|_{U}≥|p(x)|=|\delta_x(p)|$$
and $\delta_x$ extends to a character.
Not all $\delta_x$ extend to characters: If $B_r(y)$ is a ball that contains $K$ and $x$ is outside of this ball, then the polynomial $z-x$ is invertible in $\mathcal P(K)$, having inverse $\sum_n\left(\frac {z-y}{x-y}\right)^n$ (sum converges because $|x-y|=r≥|z-y|$ for all $z\in B_r(y)$). From this it follows that there can be no character with $\Phi(z)=x$.
This is as much as I know unfortunately, but I would hypothesise the following:
Hypothesis Let $U$ be the component connected to infinity of $\mathbb C-K$, if $x\in U$ then $\delta_x\notin\Phi_{\mathcal P(K)}$.
(The lemma from before shows the converse, namely that if $x\notin U$ then $\delta_x\in \Phi_{\mathcal P(K)}$).
