Clarifications the proposed result
The premise is not true in general: the enthalpy is path dependent and known only after its rate
$$
H = \int_0^t \dot U - \boldsymbol P:\dot{\boldsymbol F}\mathrm dV_0 \mathrm \;d\tau\\
\neq U +\boldsymbol P:{\boldsymbol F}\mathrm dV_0.
$$
Here we write $\mathrm dV_0=J^{-1}\mathrm dV$ the infinitesimal volume on the initial configuration.
For the given expression (with $-\boldsymbol P:{\boldsymbol F}$) to make sense in principle, we would need $\boldsymbol F$ and $\dot{\boldsymbol F}$ to remain proportional.
The only way is for $\dot{\boldsymbol F}$ to be proportional to the initial deformation gradient $\boldsymbol I$.
This corresponds to a spheric deformation, with only volume changes, which explains why only the pressure term appears at the end of OP's development.
In general there is another deviatoric term.
For the given expression to be applicable to a concrete case, one would need to find $(\boldsymbol \sigma(t),\boldsymbol F(t))$ such that, $\forall t$:
$$
\int_0^t\boldsymbol P(\boldsymbol \sigma,\boldsymbol F):\dot{\boldsymbol F}\mathrm d\tau
=\boldsymbol P:{\boldsymbol F}(t).
$$
This is tedious and far from general.
So in this case the problem was in the question formulation.
But we can still address some questions connected to the original post.
What do the Cauchy stress and its pressure term conjugate with?
Substituting $\boldsymbol P=J\boldsymbol\sigma\boldsymbol F^{-T}$ yields the power of internal forces
$$
- \boldsymbol P:\dot{\boldsymbol F}\mathrm dV_0
= - \boldsymbol\sigma:(\dot{\boldsymbol F}{\boldsymbol F}^{-1})\mathrm dV
$$
Conclusion: the Cauchy stress conjugates with the (symmetric part of the) velocity gradient $\boldsymbol l=\dot{\boldsymbol F}{\boldsymbol F}^{-1}$.
From there using $\boldsymbol\sigma=-p\boldsymbol I$ the hydrostatic part of the power of internal forces reads
$$
- \boldsymbol\sigma:\boldsymbol l\mathrm dV
= p (\mathrm{tr}\,\boldsymbol l)\mathrm dV
+ (\ldots)_\text{deviatoric}.
$$
Conclusion: the pressure conjugates with the trace of the velocity gradient.
Note: we have $\mathrm{tr}\,\boldsymbol l = \mathrm{div}\, \boldsymbol v$
and $(\mathrm{div}\, \boldsymbol v\mathrm) dV= \dot J\mathrm dV_0$
therefore $p(\mathrm{tr}\,\boldsymbol l)\mathrm dV = p\dot J\mathrm dV_0$.
The next section brings to the same result coming from a different angle.
What is obtained when the deformation is spheric?
This is to show where the missing $1/3$ would come from in a more general case.
The original post implies a deformation where only the volume changes.
This is a particular case where $\boldsymbol F$ reduces to $\boldsymbol F^\text{vol}=J^{1/3}\boldsymbol I$.
We apply chain rule to find the time derivative:
$$
\dot{\boldsymbol F}^\text{vol} = \frac13 J^{-2/3}\dot J\boldsymbol I
$$
and the corresponding velocity gradient
$$
\boldsymbol l
= \dot{\boldsymbol F}^\text{vol}\boldsymbol F^{\text{vol},-1}
= \frac13 J^{-1}\dot J
$$
The power of internal forces becomes
$$
p (\mathrm{tr}\,\boldsymbol l)\mathrm dV
= p \frac{\mathrm{tr}\,\boldsymbol I}3 \big(J^{-1}\dot J\big)\mathrm dV\\
= p \dot J\mathrm dV_0
$$
The time-and-space integration of $\dot J\mathrm dV_0$ is the total volume.
