= Solution
Use the <Minkowski metric> $g=\operatorname{diag}(1,-1,-1,-1)$. Multiplying the block <gamma matrices> and using the <Pauli matrix multiplication law> gives
$$
(\gamma^0)^2=I_4,\qquad (\gamma^j)^2=-I_4,\qquad
\{\gamma^0,\gamma^j\}=0,\qquad \{\gamma^j,\gamma^l\}=-2\delta_{jl}I_4.
$$
Thus these matrices satisfy the <Clifford algebra> relation $\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}I_4$. Since <partial derivatives> commute, the antisymmetric part of $\gamma^\mu\gamma^\nu$ drops out of its contraction with $\partial_\mu\partial_\nu$. Multiplying the <Dirac equation> on the left by $i\gamma^\mu\partial_\mu+m$ gives
$$
(i\gamma\cdot\partial+m)(i\gamma\cdot\partial-m)\psi
=-(\Box+m^2)\psi=0.
$$
Therefore \b[every component of the Dirac spinor obeys the Klein-Gordon equation with mass $m$].
For <Lorentz covariance>, take $x'=Lx$ and $\psi'(x')=S(L)\psi(x)$, with constant $L$ and $S$. The chain rule gives $\partial'_\mu=(L^{-1})^\nu{}_{\mu}\partial_\nu$. The given intertwining relation then implies
$$
\begin{aligned}
(i\gamma^\mu\partial'_\mu-m)\psi'(x')
&=S\left[i(S^{-1}\gamma^\mu S)(L^{-1})^\nu{}_{\mu}\partial_\nu-m\right]\psi(x)\\
&=S(i\gamma^\nu\partial_\nu-m)\psi(x)=0.
\end{aligned}
$$
This proves invariance of the <Dirac equation> under the joint coordinate and <spinor> transformation.
The infinitesimal identity follows directly from the <spinor Lorentz generator commutator>. To derive it, first move $\gamma^\rho$ through a two-matrix product using the <Clifford algebra>:
$$
[\gamma^\mu\gamma^\nu,\gamma^\rho]
=2g^{\nu\rho}\gamma^\mu-2g^{\mu\rho}\gamma^\nu.
$$
Subtract the expression with $\mu$ and $\nu$ interchanged and multiply by $i/2$. Thus
$$
[\sigma^{\mu\nu},\gamma^\rho]
=2i(g^{\nu\rho}\gamma^\mu-g^{\mu\rho}\gamma^\nu).
$$
Writing $S=1-\frac i4\sigma^{\mu\nu}\omega_{\mu\nu}$ gives $S^{-1}=1+\frac i4\sigma^{\mu\nu}\omega_{\mu\nu}+O(\omega^2)$, and hence
$$
\begin{aligned}
S^{-1}\gamma^\rho S
&=\gamma^\rho+\frac i4\omega_{\mu\nu}[\sigma^{\mu\nu},\gamma^\rho]+O(\omega^2)\\
&=\gamma^\rho-\frac12\omega_{\mu\nu}(g^{\nu\rho}\gamma^\mu-g^{\mu\rho}\gamma^\nu)+O(\omega^2)\\
&=\gamma^\rho+\omega^\rho{}_{\nu}\gamma^\nu+O(\omega^2)
=L^\rho{}_{\nu}\gamma^\nu+O(\omega^2).
\end{aligned}
$$
The last step uses $\omega_{\mu\nu}=-\omega_{\nu\mu}$, including the metric signs in raising the first index.
For <parity symmetry in quantum field theory>, set $y=x_P=(t,-\mathbf x)$. The time derivative of $\psi(y)$ is unchanged and each spatial derivative changes sign. Also $\gamma^0$ commutes with itself and anticommutes with every $\gamma^j$. Consequently
$$
(i\gamma^\mu\partial_\mu-m)\gamma^0\psi(x_P)
=\gamma^0(i\gamma^\mu\partial_{y^\mu}-m)\psi(y)=0.
$$
This verifies the specified <parity symmetry in quantum field theory> action explicitly.
For <charge conjugation>, put $B=i\gamma^2$, so $\psi_C=B\psi^*$. In the <Dirac representation of the gamma matrices>, $\gamma^0,\gamma^1,\gamma^3$ are real and $\gamma^2$ is imaginary. The <Clifford algebra> therefore gives $B\gamma^{\mu *}B^{-1}=-\gamma^\mu$ for all four indices. Complex conjugation of the <Dirac equation>, followed by multiplication by $B$, yields
$$
(-i\gamma^{\mu *}\partial_\mu-m)\psi^*=0
\quad\Longrightarrow\quad
(i\gamma^\mu\partial_\mu-m)\psi_C=0.
$$
This is the <antilinear charge-conjugation intertwiner> property. No electromagnetic background is being held fixed in this free-field calculation.
Apply the specified <parity symmetry in quantum field theory> action before <charge conjugation>. Since $\gamma^0$ is real and $B\gamma^0=-\gamma^0B$,
$$
(\psi_P)_C(x)=B[\gamma^0\psi(x_P)]^*
=\boxed{-\gamma^0\psi_C(x_P)}.
$$
Thus the induced parity action on the charge-conjugate spinor has the opposite sign. In particular, if a spinor has <intrinsic parity> $\eta=\pm1$, its charge conjugate has intrinsic parity $-\eta$. Equivalently, at rest the positive-frequency <Dirac spinors> occupy the upper two components and have $\gamma^0$ eigenvalue $+1$, whereas the negative-frequency <Dirac spinors> occupy the lower two components and have eigenvalue $-1$. The corresponding particle and antiparticle creation operators inherit these opposite signs under the specified parity convention. Hence \b[electrons and positrons have opposite intrinsic parities]; conventionally one assigns $+1$ to the electron and $-1$ to the positron. The common real overall parity sign may be reversed without changing this relative sign, as in <opposite intrinsic parities of a Dirac particle and antiparticle>.
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