= Solution
Put $x_P=(t,-x,y)$ and $U_P=\gamma^1$. Since $U_P^\dagger=-\gamma^1$ and $\gamma^0\gamma^1\gamma^0=-\gamma^1$, the <Dirac adjoint> transforms as
$$
\bar\psi'(x)=\psi^\dagger(x_P)(-\gamma^1)\gamma^0
=\bar\psi(x_P)\gamma^1.
$$
The <chain rule> gives $\partial_\mu\psi(x_P)=r_\mu(\partial_\mu\psi)(x_P)$, with $(r_0,r_1,r_2)=(1,-1,1)$ and no sum in this formula. The <Clifford algebra> gives
$$
\gamma^1\gamma^0\gamma^1=\gamma^0,\qquad
\gamma^1\gamma^1\gamma^1=-\gamma^1,\qquad
\gamma^1\gamma^2\gamma^1=\gamma^2.
$$
Multiplication by $r_\mu$ therefore restores each <kinetic term>:
$$
\bar\psi'i\gamma^\mu\partial_\mu\psi'
=\bar\psi(x_P)i\gamma^\mu(\partial_\mu\psi)(x_P).
$$
However, the <Dirac scalar bilinear> changes sign because $(\gamma^1)^2=-I_2$:
$$
\bar\psi'\psi'=-\bar\psi(x_P)\psi(x_P).
$$
Changing variables in the <action> has unit absolute <Jacobian determinant>. Hence
$$
\boxed{S_m\xrightarrow{P}S_{-m}:\quad S_{\rm kin}\to S_{\rm kin},\quad
-m\int\bar\psi\psi\to+m\int\bar\psi\psi.}
$$
A nonzero <Dirac mass term> for this single two-component field breaks <parity symmetry in quantum field theory> with the specified transformation.
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