So we have the commonly quoted momentum space version of the Dirac equation and the adjoint Dirac equation:

$$(\gamma^{\mu}p_{\mu}-m)u=0$$ Often, we are asked to show that the adjoint momentum Dirac equation can be written as: $$\bar{u}(\gamma^{\mu}p_{\mu}-m)=0$$ I'm not too sure on the method. However, I have attempted something.

I multiply the top equation by $\bar{u}\gamma^{\nu}\times$ and the bottom equation by $\times\gamma^{\nu}u$ giving:

$$\bar{u}\gamma^{\nu}(\gamma^{\mu}p_{\mu}-m)u=0$$

$$\bar{u}(\gamma^{\mu}p_{\mu}-m)\gamma^{\nu}u=0$$ Taking the sum of these two gave me: $$\bar{u}\gamma^{\nu}\gamma^{\mu}p_{\mu}u-\bar{u}\gamma^{\nu}mu+\bar{u}\gamma^{\mu}p_{\mu}\gamma^{\nu}u-\bar{u}m\gamma^{\nu}u=0$$ I then use $\gamma^{\mu}\gamma^{\nu}=-\gamma^{\nu}\gamma^{\mu}$, giving me: $$\bar{u}\gamma^{\nu}\gamma^{\mu}p_{\mu}u-\bar{u}\gamma^{\nu}mu-\bar{u}\gamma^{\nu}\gamma^{\mu}p_{\mu}u+\bar{u}\gamma^{\nu}mu=0$$ And which point I can say that it is true. I'm just wondering if this is sufficient or if there is another more correct way of getting the adjoint Dirac equation...

Let $$(\gamma^\mu p_\mu-m)u=0$$
Using the property $\overline{AB}=\bar{B}\bar {A}$, we have the following: $$0=\overline{(\gamma^\mu p_\mu-m)u}=\bar u \overline{(\gamma^\mu p_\mu-m)}$$
Now, use $\overline{A+B}=\bar A+\bar B$: $$0=\bar u (\overline{\gamma^\mu p_\mu}-\bar m)$$
Next, as both $m$ and $p_\mu$ are real numbers, we have $\bar m=m$ and $\bar p_\mu=p_\mu$: $$0=\bar u (\overline{\gamma^\mu} p_\mu- m)$$
Finally, use the fact that the gamma matrices are self-adjoint, that is, $\bar \gamma^\mu=\gamma^\mu$: $$0=\bar u (\gamma^\mu p_\mu- m)$$ and we are done.