Not in general. You can describe always a motion in terms of rotation around a fixed point but that won't always be a uniform rotation i.e. it might have accelerations, radial displacements etc.
What you can do is very locally approximate your motion to a rotation.
We can write the position $\vec{r}=(x(t), y(t))$ of your particle in polar coordinates (assuming motion is on a plane for simplicity) with respect to the origin as
$$\vec{r}=(\rho(t), \theta(t))$$
using the transformations
$$\rho=\sqrt{x^2+y^2}$$
and
$$\theta=atan2(y, x)$$
(see here for the definition of atan2)
You could also do the same for any point $P=(x_0, y_0)$ substituing $x\to x-x_0$ and $y\to y-y_0$.
Then of course $$\omega=d\theta/dt$$ but you already see that this involves a derivative of the atan2 function so it is not in general going to be a constant speed as you ask in your question.
Now however, if your displacement is really small you could locally approximate it to a rotation $\hat{R}$ so that $$\vec{r}(t+dt)=\hat{R}\vec{r}$$
where $\hat{R}$ is a rotation matrix.
If the displacement is very small, you could use the infinitesimal form of the rotation (assuming $P$ is the origin) $$\hat{R}=\begin{bmatrix}cos(d\theta) & -sin(d\theta) \\ sin(d\theta) & cos(d\theta)\end{bmatrix}\approx\begin{bmatrix}1 & -d\theta\\d\theta & 1\end{bmatrix}$$ where I used $cos(d\theta)\approx 1$ and $sin(d\theta)\approx d\theta$ for very small angles, so that, applying the matrix
$$\vec{r}(t+dt)=\hat{R}\vec{r}$$
becomes
$$\vec{r}(t+dt)=\begin{bmatrix}x(t+dt) \\y(t+dt)\end{bmatrix}=\begin{bmatrix}1 & -d\theta\\d\theta & 1\end{bmatrix}\begin{bmatrix}x(t) \\y(t)\end{bmatrix}$$
which we can separate in two equations
$$x(t+dt)=x(t)-y(t)d\theta$$
$$y(t+dt)=x(t)d\theta+y(t)$$
which however assumes the distance from the origin is constant (so you have an error in the radial direction is $\rho$ is actually changing).
If we rewrite that as
$$dx=x(t+dt)-x(t)=-y(t)d\theta$$
$$dy=y(t+dt)-y(t)=x(t)d\theta$$
you get that
$$dr^2=dx^2+dy^2=d\theta^2(y^2+x^2)=d\theta^2\rho^2(t)$$
meaning
$$d\theta=dr/\rho(t)$$
so that this whole thing is valid if
$$d\theta\ll 1$$
i.e.
$$dr\ll \rho(t)$$
meaning that you can do it only if your total displacement $dr$ is much smaller than the radius of the circle with which you are approximating your motion. Also, this is valid at one timestep $dt$ and the more time-steps you do, the less this approximation is going to be valid if your radius changes consistently over time.
Now, in terms of speed, becuase $dr=vdt$ we get
$$d\theta=dr/\rho(t)=vdt/\rho(t)$$ so that
$$\omega = d\theta/dt = v / \rho$$ as expected, but again: it is only valid for an infinitesimal displacement. As you move away from the circle or even if you rotate around the origin at non-constant angular acceleration, this approximation sooner or later fails. Especially, as mentioned, you are completely neglecting displacement in the radial position, so if your motion is completely radial this is completely wrong as $d\theta = 0$ meaning you would get $\vec{r}(t+dt)=\hat{R}\vec{r}(t)=\vec{r}(t)$ i.e. you are not moving at all.