A correspondent suggests that "Your question is not a noncommutative issue. It suffices to look at the restriction on the commutative $\mathrm{C}^*$-subalgebra generated by $f$ and by $\pi(f)$ and then everything is clear."
Following this line define $\mathcal{A}_0:=\mathrm{C}^*(f)\cong C([a,b])$, and, where $\Lambda=\sigma(\pi(f))=\{\lambda_1,\dots,\lambda_m\}$, define $\mathcal{B}_0:=\mathrm{C}^*(\pi(f))\cong C(\Lambda)$. In $\mathcal{A}_0$, $f(z)=z$ and $\pi(f)=\sum_{\lambda\in\Lambda}\lambda\,\delta_\lambda$. For $g\in \mathcal{A}_0$, $\pi(g)=\sum_{\lambda\in\Lambda}g(\lambda)\delta_\lambda$.
By assumption, implicitly using restrictions, for some $\lambda\in\Lambda$, $\varphi_0(\delta_\lambda)=1$ and so
$$\varphi(g)=\varphi_0\left(\sum_{\lambda\in\Lambda}g(\lambda)\delta_\lambda\right)=g(\lambda)\Rightarrow \varphi=\operatorname{ev}_{\lambda}.$$
The enveloping von Neumann algebra $\mathcal{A}_0^{**}\cong \ell^{\infty}([a,b])\subset B(\ell^2([a,b]))$ and $\varphi$ extends to $\omega_\varphi=\operatorname{ev}_{\lambda}\in \ell^{\infty}([a,b])^*$. The spectral projection $\mathbf{1}_{\{\lambda\}}(f)$ is:
$$\ell^2([a,b])\ni \phi\mapsto \phi(\lambda)\delta_{\lambda}\in\ell^2([a,b]),$$
so that $\mathbf{1}_{\{\lambda\}}(f)$ may be identified with $\delta_{\lambda}\in\ell^{\infty}([a,b])$ and indeed $\omega_\varphi(\mathbf{1}_{\{\lambda\}}(f))=\operatorname{ev}_{\lambda}(\delta_{\lambda})=1$, as required.
I think the answer is again no. Take $A=B=\mathbb C^2$, so that $C=A\otimes B = \mathbb C^4$.
The goal is to reduce this to my answer to your previous question by choosing an appropriate state.
Let $e_1$ and $e_2$ be the canonical basis vectors in $\mathbb C^2$, and choose the state $\phi$ on $C$ such that $\phi(e_i\otimes e_j)=1/3$, for all $i,j$, except for $\phi(e_2\otimes e_1)=0$.
If $(\pi,H,\Omega)$ is the corresponding GNS representation we then have that
$\pi(e_2\otimes e_1)\Omega =0$, and hence
$$
f_1:= \pi (e_1\otimes e_1)\Omega ,
$$
$$
\qquad\ f_2:= \pi (e_1\otimes e_2)\Omega ,\quad \text{and}
$$
$$
f_3:= \pi (e_2\otimes e_2)\Omega
$$
span $H$.
It is not hard to see that these form a linearly independent set, and hence that $H$ is 3-dimensional.
Given any $a$ in $A$, say $a=(\lambda ,\mu )$, we have that
$$
\pi (a\otimes 1) = \pi ((\lambda e_1 + \mu e_2)\otimes 1)= \lambda \pi (e_1\otimes 1) + \mu \pi (e_2\otimes 1),
$$
so it is easy to see that
$$
\pi (a\otimes 1)f_1 = \lambda f_1, \quad \pi (a\otimes 1)f_2 = \lambda f_2, \quad \text {and} \quad \pi (a\otimes 1)f_3 = \mu f_3,
$$
whence the matrix of $\pi (a\otimes 1)$ in the basis $\{f_1, f_2, f_3\}$ is
$$
\pmatrix{\lambda & 0 & 0 \cr 0 & \lambda & 0 \cr 0 & 0 & \mu }.
$$
If instead $b = (\lambda ,\mu )\in B$, then
$$
\pi (1\otimes b) = \pi (1\otimes (\lambda e_1 + \mu e_2)) =\lambda \pi (1\otimes e_1) + \mu \pi (1\otimes e_2),
$$
so
a similar analysis gives the matrix of $\pi (1\otimes b)$, as being
$$
\pmatrix{\lambda & 0 & 0 \cr 0 & \mu & 0 \cr 0 & 0 & \mu }.
$$
Identifying $H$ with ${\mathbb C}^3$ via the basis mentioned above we see that $\Omega$ is proportional to $(1,1,1)$, and
$$
[\pi (A\otimes 1)\Omega ] = \{(\lambda , \lambda , \mu ): \lambda , \mu \in {\mathbb C}\},
$$
and
$$
[\pi (1\otimes B)\Omega ] = \{(\lambda , \mu , \mu ): \lambda , \mu \in {\mathbb C}\},
$$
whence
$$
P_A =\pmatrix {1/2 & 1/2 & 0 \cr 1/2 & 1/2 & 0 \cr 0 & 0 & 1},
$$
and
$$
P_B =\pmatrix { 1& 0 & 0 \cr 0 & 1/2 & 1/2 \cr 0 & 1/2 & 1/2 },
$$
which do not commute.
I am still thinking about what happens when $\phi$ is a product state and I believe the answer is affirmative in that case.
Best Answer
Recall that for a linear map $T:X\to Y$ between normed spaces, we define its adjoint $T^*:Y^*\to X^*$ as $T^*(\phi)=\phi\circ T$. We have that $\|T^*\|=\|T\|$.
The answer is that we can always extend $*$-homomorphisms on the double duals: given a $*$-homomorphism $\varphi:A\to B$ between C*-algebras, the double adjoint map $\varphi^{**}:A^{**}\to B^{**}$ is a $*$-homomorphism on the double duals that extends $\varphi$. Moreover, $\varphi^{**}$ is ultraweakly continuous. Moreover, if $\varphi$ is injective, then so is $\varphi^{**}$.
The fact that $\varphi^{**}$ extends $\varphi$ and that $\varphi^{**}$ is ultraweakly continuous is obvious. I will show for example that $\varphi^{**}$ preserves multiplication: We have that $A$ is ultraweakly dense in $A^{**}$ by Goldstine's theorem. Let $x,y\in A^{**}$ and find norm bounded nets $(x_i),(y_i)$ in $A$ such that $x_i\to x$ and $y_i\to y$ ultraweakly. Fix an index $i_0$. Using in the 1st and last equation below the fact that multiplication is separately continuous for the ultraweak topology on bounded sets, we have that $$\varphi^{**}(xy_{i_0})=\lim_i\varphi^{**}(x_iy_{i_0})=\lim_i\varphi(x_iy_{i_0})=\lim_i\varphi(x_i)\varphi(y_{i_0})=\varphi^{**}(x)\varphi(y_{i_0})$$ so $\varphi^{**}(xy_{i_0})=\varphi^{**}(x)\varphi^{**}(y_{i_0})$. As this is true for all indices $i_0$, taking ultraweak limits and employing the same argument yields that $\varphi^{**}$ preserves multiplication. A similar (less complicated) argument shows that $\varphi^{**}$ preserves involution.
The claim that $\varphi$ injective implies $\varphi^{**}$ injective is true more generally for linear maps between normed spaces that have closed image (an application of Hahn-Banach is needed).
With some extra work, one can also show that if $\varphi:A\to B$ is a completely positive map, then $\varphi^{**}:A^{**}\to B^{**}$ is also completely positive.