Spatial inversion and Time reversal On the spinor field $\psi^{\mu}(x)$, I found the action of $\psi^{\mu}(x)$ on spatial inversion $P$ by postulating $\psi^{\mu}_{P}(x)=P^{\mu}_{\nu}\psi^{\nu}(P^{-1}x)=P^{\mu}_{\nu}\psi^{\nu}(t,-x)$, is $P=\pm \gamma^0$ , Thus $$\psi^{\mu}_{P}(x)=\pm \gamma^0\psi^{\mu}(t,-x)$$
Now, my question is how do the action of vector field $U^{\mu}$ on spatial inversion $P$ will look like?
Any useful comments/answer are welcome.
 A: Quoting this answer of mine, you cannot derive how a field transforms, under any transformation. All you can do is make definitions.
There is no general way a field should transform under parity (or any other symmetry). There are situations where a certain definition will be useful, and other situations where a different definition is required. The point is: the "correct" transformation is whatever makes the Lagrangian invariant. This depends, of course, on the specific Lagrangian you are working with, and there is no one-size-fits-all answer.
If your Lagrangian contains the standard Yang-Mills term, then the following definition is a good symmetry:
$$
(\mathcal P\cdot U)^\mu(t,x):=\eta^{\mu\mu}U^\mu(t,-x)\tag1
$$
where $\eta$ is the Minkowski metric and the summation convention is on pause. (Mostly minus for concreteness, but mostly plus works too).
This definition is what one may call the "fundamental" (or "vector/standard/defining") representation, because the vector field $U^\mu$ transforms in the same way as the vector $x^\mu$. In more precise terms, $U^\mu$ was defined to transform as the fundamental rep of $SO(1,d)$, and now we are claiming that it in fact lifts to the fundamental rep of $O(1,d)$.
But, and I cannot stress this enough, this definition is by no means unique. You may very well be dealing with an exotic QFT where the kinetic term is not the standard Yang-Mills term, or the interactions mix the components in a weird way, in which case the definition $(1)$ may no longer be a symmetry of your Lagrangian. In such a case, you will need to come up with a different definition which does work. And I cannot tell you the answer because the answer is extremely non-unique.
A: Your question is not very clear to me, but I'm assuming that you mean to ask what is the action of the parity operator on a vector field? (By this I assume you mean, something born out of the tensorial representation of the proper Lorentz group: Since, you have already talked about the Dirac representation)

Along similar lines,
$$
V^{\mu}(x) \rightarrow \mathsf{P}V^{\mu}(\mathcal{P}x) = \mathsf{P}^{\mu}{}_{\nu}V^{\nu}(\mathcal{P}x)
$$
But, we already know what $\mathsf{P}$ is, since we are working with the tensorial representation.
$$
\mathsf{P} = \begin{pmatrix}
\boldsymbol{t} & \boldsymbol{x^{1}} & \boldsymbol{x^{2}} & \boldsymbol{x^{3}} \\
1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & -1 \\
\end{pmatrix}
$$
