Answer to original, pre-edit question:
No, because in general $p$ need not be an upper bound for that set.
For example, suppose $A$ is the image of the universal representation of the algebra of 2-by-2 complex matrices $M_2$, and to hopefully reduce confusion I want to explicitly mention an isomorphism $\phi:M_2\to A$. Consider the vector state $\tau$ on $M_2$ induced by $(0,1)\in\mathbb{C}^2$ and let $\mu = \tau\circ\phi^{-1}$ be the corresponding state on $A$. Then $a=\phi\left(\left( \begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix} \right)\right)\in A$ satisfies $\mu(a)=0$ and $0\leq a\leq1$. However, $a\nleq0$, and $p$ in this case is 0 because $\mathbf{ker} \ \pi_\mu'' = \mathbf{ker} \ \pi_\mu = \{0\}$.
Addressing new question:
As you indicated, the problem with the original question is that $\mu(a)=0$ does not typically imply that $\pi_\mu(a)=0$. Your new condition does the tricktakes care of this: If $x_\mu\in\mathcal{H}_\mu$ is the GNS vector for $\pi_\mu$ and $\pi_\mu(a)\neq0$, then because the numerical radius of $\pi_\mu(a)$ is nonzero and $x_\mu$ is cyclic for $\pi_\mu$, there is a $c\in A$ such that $\langle\pi_\mu(a)(\pi_\mu(c)x_\mu),\pi_\mu(c)x_\mu\rangle\neq0.$ That is, $\mu(c^*ac)\neq0$.
(The question thus reduces to whether $p=\sup\{a\in\mathbf{ker} \ \pi_\mu: 0\leq a\leq1\}$, and this holds thus by Kaplansky's density theorem to whether $\mathbf{ker} \ \pi_\mu$ is dense in $\mathbf{ker} \ \pi_\mu''$. I don't have an argument for why that is true.)
