Maybe, the answer will be non-useful for the author of the question, but it is important for the other people who want to derive this expression. [:)].
First method.
I made some hint: it's easy to show, that
$$
\hat {W}^{\mu}\hat {P}_{\mu} = \frac{1}{2}\varepsilon^{\mu \alpha \beta \gamma}\hat {J}_{\alpha \beta}\hat {P}_{\gamma }\hat {P}_{\mu} = 0
$$
as the convolution of symmetrical $\hat {P}_{\gamma }\hat {P}_{\mu}$ and antisymmetrical $\varepsilon^{\mu \alpha \beta \gamma}$.
So the commutator
$$
[\hat {J}_{\kappa \lambda}, \hat {W}^{\mu}\hat {P}_{\mu}] = 0. \qquad (.1)
$$
But by the other hand
$$
[ \hat {J}_{\kappa \lambda}, \hat {W}^{\mu}\hat {P}_{\mu}] = \hat {W}^{\mu}[ \hat {J}_{\kappa \lambda}, \hat {P}_{\mu}] + [\hat {J}_{\kappa \lambda }, \hat {W}^{\mu}]\hat {P}_{\mu} = -\hat {W}^{\mu}i(g_{\mu \kappa }\hat {P}_{\lambda} - g_{\mu \lambda}\hat {P}_{\kappa }) + [\hat {J}_{\kappa \lambda }, \hat {W}^{\mu}]\hat {P}_{\mu} .
$$
So, by using $(.1)$ from the previous formula there follows the next:
$$
[\hat {J}_{\kappa \lambda }, \hat {W}^{\mu}]\hat {P}_{\mu} = i\hat {W}^{\mu}(g_{\mu \kappa }\hat {P}_{\lambda} - g_{\mu \lambda}\hat {P}_{\kappa }) = i(\hat {W}_{\kappa}\hat {P}_{\lambda} - \hat {W}_{\lambda}\hat {P}_{\kappa}) = i\left(\hat {W}_{\kappa}\delta^{\mu}_{\lambda} - \hat {W}_{\lambda}\delta^{\mu}_{\kappa}\right)\hat {P}_{\mu},
$$
where $\delta^{0}_{0} = 1, \delta^{i}_{i} = -1$,
and, finally,
$$
[\hat {J}_{\kappa \lambda }, \hat {W}^{\mu}] = i\left(\hat {W}_{\kappa}\delta^{\nu}_{\lambda} - \hat {W}_{\lambda}\delta^{\mu}_{\kappa}\right) \Rightarrow [\hat {J}_{\kappa \lambda }, \hat {W}_{\mu}] = i\left(\hat {W}_{\kappa}g_{\mu \lambda} - \hat {W}_{\lambda}g_{\mu \kappa}\right).
$$
Second method.
It is similar to QMechanic method. Lets have general Poincare transformation matrix $U (\hat {\Lambda}, a^{\mu})$, where $a$ is translation 4-vector and $\hat {\Lambda}$ is a 3-rotations and Lorentz transformations matrix. Due to this transformation,
$$
\hat {J}_{\mu \nu} \Rightarrow U (\hat {\Lambda}, a^{\mu}) J^{\mu \nu}U (\hat {\Lambda}, a^{\mu})^{+} = \Lambda^{\mu}_{\rho}\Lambda^{\nu}_{\sigma}\hat {J}^{\rho \sigma} + a^{\mu}\Lambda^{\nu}_{\rho }\hat {P}^{\rho} - a^{\nu}\Lambda^{\mu}_{\rho}\hat {P}^{\rho },
$$
$$
\hat {P}^{\mu} \Rightarrow \Lambda^{\mu}_{\nu}\hat {P}^{\nu}.
$$
By using these transformation rules you can get
$$
U (\hat {\Lambda}, a) W_{\mu} U (\hat {\Lambda}, a)^{+} = \frac{1}{2}
\varepsilon_{\mu \nu \rho \sigma}\left( \lambda^{\nu}_{\alpha}\Lambda^{\rho}_{\beta}\hat {J}^{\alpha \beta} + a^{\nu}\Lambda^{\rho}_{\alpha }\hat {P}^{\alpha} - a^{\rho}\Lambda^{\nu}_{\rho}\hat {P}^{\rho }\right) \Lambda^{\sigma}_{\delta}\hat {P}^{\delta } = $$
$$
=\frac{1}{2}\Lambda^{\alpha}_{\mu}\varepsilon_{\alpha \nu \rho \sigma }\hat {J}^{\nu \rho} \hat {P}^{\sigma} = \Lambda^{\alpha}_{\mu} \hat {W}_{\alpha}.
$$
So
$$
\frac{i}{2}[\hat {J}_{\mu \nu}, \hat {W}_{\sigma }] = \omega^{\mu \nu}g_{\sigma \mu}\hat {W}_{\nu} = \frac{1}{2}(g_{\sigma \mu}\hat {W}_{\nu} - g_{\sigma \nu}\hat {W}_{\mu})\omega^{\mu \nu},
$$
which leads to the targeted expression.
In general, this means that all of 4-operator $\hat {A}_{\gamma}$ commutes with $\hat {J}_{\alpha \beta}$ as $i(g_{\gamma \alpha}\hat {A}_{\beta} - g_{\gamma \beta}\hat {A}_{\alpha})$, because $\hat {J}_{\alpha \beta}$ represents the generators of the Lorentz group. Also, this means, that commutator of $\hat {J}_{\alpha \beta }$ with each 4-scalar will give zero.