To prove equation 9.91 globally we will first prove it locally, the generalization is then straightforward.
We take a point $p$ of the spacetime-manifold $M$, at which a tangent space $T_p$ is defined.
The scalar fields $\phi(x^{\mu})$ are then defined with respect to a coordinate system constructed from the basis vectors of $T_p$.
Now, in general relativity, curved spacetime looks locally like Minkowski space + a gravitational force (Equivalence Principle). So we can construct Riemann Normal Coordinates $x^{\hat{\mu}}(p)$ satisfying:
$$g_{\hat{\mu} \hat{\nu}}(p) = \eta_{\hat{\mu} \hat{\nu}}, \partial_\hat{\sigma} g_{\hat{\mu} \hat{\nu}}(p) = 0.$$
These coordinates are called locally inertial coordinates (see eq 2.47 in Carroll's book).
Next relabel $g_{\hat{\mu} \hat{\nu}}$ to $g_{\mu\nu}$ to avoid confusion. Then, we show that equation 9.91 holds in these coordinates:
$$\pi = \frac{\partial}{\partial(\nabla_0 \phi)} ( \sqrt{-g} \{-\frac{1}{2} g^{\mu\nu} \nabla_\mu \phi \nabla_\nu \phi - \cdots \})$$
$$ = \frac{\partial}{\partial(\nabla_0 \phi)} ( \sqrt{-g} \{-\frac{1}{2} g^{00} \nabla_0 \phi \nabla_0 \phi + g^{0i} \nabla_0 \phi \nabla_i \phi + g^{i0} \nabla_i \phi \nabla_0 \phi + g^{ij} \nabla_i \phi \nabla_j \phi - \cdots \})$$
by only looking at the first term, as the rest does not depend on $\nabla_0 \phi$, we further derive:
$$ \frac{\partial}{\partial(\nabla_0 \phi)} (g^{00} \nabla_0 \phi \nabla_0 \phi) = 2 g^{00} \nabla_0 \phi;$$
just use the Leibniz rule to prove this statement. Then:
$$\pi = \sqrt{-g} \nabla_0 \phi.$$
Where the metric is put in its canonical form $g_{\mu \nu} = diag(-1, +1, +1, +1)$ as we are using locally inertial coordinates.
Finally, as $\pi = \sqrt{-g} \nabla_0 \phi$ is a tensorial equation (as the covariant derivative of a scalar field is independent of the used coordinate system) so 9.91 is globally true.
It should be noted that $g^{00} = g_{00} = -1$ and by this equation 9.91 is true.
