The pressure is solved for. That said, it is common to apply the Leray projector $\mathbb P$ (projection to the space of divergence free vector fields; recall the Helmholtz decomposition) to the Navier-Stokes equation to obtain an equation without explicit mention of $p$,
$$ \partial_t u + \mathbb P ((u\cdot\nabla) u) - \nu \Delta u = \mathbb P g$$
If you manage to find the solution $u$ to this system, then $p$ can be recovered (assuming everything is $C^\infty$ and integrable enough, say) by taking the divergence of the Navier-Stokes equation, leaving you with a Poisson equation for the pressure $p$:
$$ -\frac1{\rho_0}\Delta p = \nabla\cdot ((u\cdot\nabla u)) - \nabla \cdot g $$
(the density $\rho \equiv \rho_0$ is constant for incompressible N-S) so you can use the solution theory for Poisson equations to find $p$.
I learned this from the book of Majda and Bertozzi "Vorticity and Incompressible Flow", and I believe it is also in this more recent book "The Three-Dimensional Navier–Stokes Equations" by Robinson, Rodrigo, & Sadowski. Its got to be in most books on the incompresible Navier-Stokes equation.
This really depends on the nature of the domain $D$. If the domain $D(t)$ is time-dependent where $\mathbf{u}^{(b)}(\mathbf{x})$ is the velocity of an area element at the boundary point $\mathbf{x} \in \partial D(t)$, then according to Reynolds transport theorem (proved here) and an application of the divergence theorem, we have
$$\tag{1}\frac{dK}{dt} = \int_{D(t)} \frac{\partial K}{\partial t}\, dV + \int_{ D(t)}\nabla \cdot \left(K\mathbf{u}^{(b)}\right) \, dV \\= \int_{D(t)} \frac{\partial K}{\partial t}\, dV + \int_{\partial D(t)}K\left(\mathbf{u}^{(b)}\cdot \mathbf{n}\right) \, dA$$
There are three possibilities.
(i) We have an arbitrary moving domain with some prescribed boundary velocity $\mathbf{u}^{(b)}$ and (1) cannot be simplified further without more information.
(ii) We have a solid stationary boundary where $\mathbf{u}^{(b)} \cdot \mathbf{n}$ holds at every boundary point either as a consequence of the no-slip condition for a viscous fluid or the impermeability condition for an inviscid fluid. In this case (1) reduces to
$$\tag{2}\frac{dK}{dt} = \int_{D(t)} \frac{\partial K}{\partial t}\, dV $$
(iii) The domain is a material element where $\mathbf{u}^{(b)}= \mathbf{u}$, that is
the velocity at the boundary is the fluid velocity. In this case,
$$\tag{3}\frac{dK}{dt} = \int_{D(t)} \frac{\partial K}{\partial t}\, dV + \int_{\partial D(t)}K\left(\mathbf{u}\cdot \mathbf{n}\right) \, dA$$
Again, the ambiguity of $\mathbf{u} \cdot \mathbf{n}$ remains.
In the above discussion nothing was said about the incompressiblity of the fluid and the condition $\nabla \cdot \mathbf{u} = 0$, because it was not necessary to derive either form of the Reynolds transport theorem shown in (1).
What you show in your first equation
$$\frac{d}{dt} K(t) = \int_D \rho \dot{u} \cdot u \, dV,$$
is a consequence of the Reynolds transport theorem given both incompressiblity and the specific form $K = \frac{1}{2} \rho |\mathbf{u}|^2$. This follows form
$$ \frac{\partial K}{\partial t} +\nabla \cdot (K\mathbf{u}) = \frac{\partial K}{\partial t} + \underbrace{K \nabla \cdot \mathbf{u}}_{= 0} + (\mathbf{u} \cdot\nabla)K \\ = \frac{\rho}{2} \frac{\partial }{\partial t} (\mathbf{u} \cdot \mathbf{u}) + \frac{\rho}{2}(\mathbf{u} \cdot \nabla) (\mathbf{u} \cdot \mathbf{u}) = \rho \frac{\partial \mathbf{u}}{\partial t} \cdot \mathbf{u} + \rho \left[(\mathbf{u} \cdot \nabla) \mathbf{u} \right] \cdot \mathbf{u} \\ := \rho \dot{u}\cdot u$$
However, this special form does not make it any clearer why the two forms you present should be equal in all circumstances.
Best Answer
Let us consider the initial-value problem $\rho(x,0) = \rho_0(x)$. The method of characteristics consists in introducing the parameterization $\rho(x(t),t)$ of the density. Differentiation w.r.t. time gives $$ \frac{\text d }{\text d t}\rho = \rho_x\, \frac{\text d }{\text d t}x + \rho_t \, . $$ Using the PDE $\rho_t + u \rho_x = 0$, one writes the system of characteristic equations $$ \frac{\text d }{\text d t}\rho = 0\, , \qquad \frac{\text d }{\text d t}x = u\, . $$ Letting $\rho(0) = \rho_0(x_0)$ gives $\rho(x(t),t) = \rho_0(x_0)$, and letting $x(0) = x_0$ gives $x(t) = x_0 + \int_{0}^{t}u(s)\,\text d s$. Combining both equations and going back to the original variables, we have $$ \rho(x,t) = \rho_0\left(x - \int_{0}^{t}u(s)\,\text d s\right) . $$ The constant velocity case $u(t) = \bar u$ is included in this formula, which gives $\rho(x,t) = \rho_0(x - \bar u t)$.