Put simply, steady-state is a pointwise phenomenon, not a global system phenomenon. To answer your question, I will provide an example of a steady state system for which $\partial p / \partial t = 0$ but $d p / d t \neq = 0.$
Think of a horizontal stream of fluid on the $x$-axis flowing in the $+x$ direction. Suppose the velocity of the fluid may change at different points along the stream, as may the stream's width, but at any given point along the $x$ axis, the velocity and the width remain constant with time. This is a steady state system—if I take a picture of the stream at time $t = 0s$ and take another picture at time $t = 10 s$, the two pictures will look identical. In this time interval a great volume of fluid may have passed through each point in the stream, but the system as a whole looks the same as it did ten seconds in the past.
Let $v(x, t)$ be the velocity of the stream at a given $x$ position, and $w(x, t)$ be its width. The loose notion of "steady state" we gave above is put more rigorously as follows:
This stream is in a steady state if at any given point $x$ in the stream, the quantities of interest $w(x, t)$ and $v(x, t)$ are unchanging with time.
Since we are fixing a point $x$ in the stream, the above is equivalent to demanding $\frac{\partial v}{\partial t} = \frac{\partial w}{\partial t} = 0.$
The total derivatives may not be zero anywhere. For example, we have
$$
\frac{d v}{d t} = \frac{\partial v}{\partial x} \frac{d x}{d t} + \frac{\partial v}{\partial t}.
$$
If the velocity gradient $\partial v / \partial x$ is nonzero, and the velocity $d x / d t$ at a given point is nonzero, then the total derivative $dv / dt$ is nonzero. That is, if I look at a single particle of the fluid, of course its velocity changes with time. It moves along the stream, and the velocity changes at different points in the stream.
But at any given point, the velocity of all particles passing through that point is constant for all time. This is what is meant by "steady state."
$\nabla\cdot \mathbf{A} = 0$ is known as the Coulomb (or transverse) gauge. It's also called the radiation gauge. Note that, for boundary conditions at infinity, the Coulomb gauge is complete - there is no ambiguity left for making further modifications to the gauge fields that don't violate $\lim_{r\rightarrow \infty} A^\mu(r, t) = 0$ (because any function that satisfies $\nabla^2 f(\mathbf{x})=0$ and $\lim_{r\rightarrow \infty} f(\mathbf{x}) = 0$ is identically zero).
$\phi = 0$ is a different gauge called the Weyl gauge. The residual gauge symmetries aren't enough to make it compatible with the Coulomb gauge. This gauge is unspoilt by any gauge transformation that satisfies $\frac{\partial f}{\partial t} = 0$, leaving
$$\mathbf{A}'(\mathbf{x}, t) = \mathbf{A}(\mathbf{x}, t) + \nabla f(\mathbf{x}).$$
True, it might be possible to use the residual freedom in the Weyl gauge to satisfy the Coulomb gauge condition at some particular time, but it will only be satisfied for that instant, and not in general.
Best Answer
Clearly, you don't understand this equations in a certain sense. Equation of continuity is suitable for any situation, so whenever you can write down$$\nabla \cdot\vec{J}=-\frac{\partial \rho}{\partial t}$$ Together with Ohm's law $\vec{J}=\sigma\vec{E}$, we have$$\nabla \cdot\vec{J}=\sigma\nabla \vec{E}=-\frac{\partial \rho}{\partial t}$$ So far, this goes well. For Steady-state curent, $-\frac{\partial \rho}{\partial t}\approx 0$, so $$\nabla \cdot\vec{J}=\sigma\nabla \vec{E}=0$$ In steady-state curent, there doesn't exist static charges or no accumulated charges, so $\nabla\cdot\vec{E}=0$. So it is self-consistent.