Dirac spinors under Parity transformation or what do the Weyl spinors in a Dirac spinor really stand for?

You are looking for a unitary representation of parity on spinors. That it should be unitary can be seen from the fact, that parity commutes with the Hamiltonian. Compare this to time-reversal and charge conjugation, which anticommute with $P^0$ and hence need be antiunitary and antilinear. They involve complex conjugation.

As demonstrated parity transforms a $(\frac{1}{2},0)$ into a $(0,\frac{1}{2})$ representation. Hence it cannot act on any such representation alone in a meaningful way. The Dirac-spinors in the Weyl-Basis on the other hand contain a left- and right-handed component $$ \Psi = \begin{pmatrix} \chi_L \\ \xi_R \end{pmatrix} $$

As a linear operator on those spinors - a matrix in a chosen basis - it mixes up the spinor components. After what has been said before, left- and right-handed components should transform into each other. The only matrix one can write down that does this is $\gamma^0$. There could in principle be a phase factor. In a theory with global $U(1)$-symmetry this may be set to one however.

Edit: Statements like $\chi_L \rightarrow P\chi_L=\chi_R $ for a Weyl-Spinor $\chi_L$ are not sensible. The Weyl-Spinors are reps. of $\mathrm{Spin(1,3)}$, whereas $P\in \mathrm{Pin(1,3)}$. One cannot expect that some representation is also a representation of a larger group. Dirac-Spinors on the other hand are precisely irreps. of $\mathrm{Spin(1,3)}$ including parity, which cannot act in any other sensible way than by exchanging the chiral components.

Think of what representation means. It's a homomorphism from a group to the invertible linear maps on a vectorspace. $$ \rho: G \rightarrow GL(V)$$ Particulary, for any $g\in G$ and $v\in V$, $\rho(g)v\in V$. Now set $V$ to be the space of say left-handed Weyl-Spinors and $g=P\in\mathrm{Pin(1,3)}$ the parity operation. As you have shown above, the image of a potential $\rho(P)$ is not a left-handed Weyl-Spinor, hence is not represented.

I find things are clearer using the dotted and undotted spinor notation. The L-spinors $\chi_{L}$ are dotted vectors $\chi^{\dot{A}}$ and the R-spinors $\xi_{R}$ are undotted vectors $\xi^{A}$ with index $A=1,2$. The parity operator has to be a tensor $P^{\dot{A}}_{B}$ and another tensor $P^{A}_{\dot{B}}$ in order to change the way each type of spinor transforms. The action of parity on $\chi^{\dot{A}}$ is to make $P^{A}_{\dot{B}}\chi^{\dot{B}}$ which transforms as an undotted spinor. Similarly, the action of parity on $\xi^{A}$ is to make $P^{\dot{A}}_{B}\xi^{B}$ which transforms as a dotted spinor. It turns out that (presumably in the rest frame of the particles) the parity tensors are $P^{\dot{A}}_{B}=i\delta^{\dot{A}}_{B}$ and $P^{A}_{\dot{B}}=i\delta^{A}_{\dot{B}}$. The action of parity is then, $$ \chi^{\dot{A}}\rightarrow P^{A}_{\dot{B}}\chi^{\dot{B}}=i\delta^{A}_{\dot{B}}\chi^{\dot{B}}=i\chi^{A} $$ $$ \xi^{A}\rightarrow P^{\dot{A}}_{B}\xi^{B}=i\delta^{\dot{A}}_{B}\xi^{B}=i\xi^{\dot{A}} $$ and this means that the components of the spinors get a phase and the way the components transform is changed. The dots are a reminder of how each component transforms. The action of parity on a Dirac spinor is obtained from the above transformations by stacking the Weyl spinors.

\begin{equation} \left[ \begin{array}{c} \chi^{\dot{1}}\\ \chi^{\dot{2}}\\ \xi^{1}\\ \xi^{2} \end{array} \right] \rightarrow i\left[ \begin{array}{c} \xi^{\dot{1}}\\ \xi^{\dot{2}}\\ \chi^{1}\\ \chi^{2} \end{array} \right] \end{equation}

Edit : Clarification. The Dirac spinor has four components. Components one and two transform as the two components of a dotted Weyl spinor and components three and four transform as the components of an undotted Weyl spinor. If we remember that is how the Weyl spinors stack into a Dirac spinor then we can remove the dots and the L and R labels and then the last equation on the action of parity on a Dirac spinor is, \begin{equation} \left[ \begin{array}{c} \chi^{1}\\ \chi^{2}\\ \xi^{1}\\ \xi^{2} \end{array} \right] \rightarrow i\left[ \begin{array}{c} \xi^{1}\\ \xi^{2}\\ \chi^{1}\\ \chi^{2} \end{array} \right] \end{equation} In matrix notation this is, \begin{equation} \left[ \begin{array}{cc} i & 0\\ 0 & i \end{array} \right] \left[ \begin{array}{c} \chi\\ \xi \end{array} \right]=i\left[ \begin{array}{c} \xi\\ \chi \end{array} \right] \end{equation} Modulo the phase factor $i$ (because I think that the parity operator on spinors has to be PP=-1), this agrees with the action of parity given by the gamma matrix $\gamma_{0}$ as in the standard texts such as equation (1.4.42) on page 19 of Ramond's "Field Theory", Second Edition.