With all the constants made explicit...
$$\begin{align}
\mathcal{L}
&= iħ \bar{ψ} γ^μ ∂_μ ψ - mc \bar{ψ} ψ\\
Δ\mathcal{L}
&= iħ Δ\bar{ψ} γ^μ ∂_μ ψ + iħ \bar{ψ} γ^μ ∂_μ Δψ - mc Δ\bar{ψ} ψ - mc \bar{ψ} Δψ\\
&= iħ Δ\bar{ψ} γ^μ ∂_μ ψ + \left(∂_μ (iħ \bar{ψ} γ^μ Δψ) - iħ ∂_μ\bar{ψ} γ^μ Δψ\right) - mc Δ\bar{ψ} ψ - mc \bar{ψ} Δψ\\
&= Δ\bar{ψ} \left(iħ γ^μ ∂_μ ψ - mc ψ\right) - \left(iħ ∂_μ\bar{ψ} γ^μ + mc \bar{ψ}\right) Δψ + ∂_μ \left(iħ \bar{ψ} γ^μ Δψ\right)\\
&⇒ iħ γ^μ ∂_μ ψ - mc ψ = 0, \quad iħ ∂_μ\bar{ψ} γ^μ + mc \bar{ψ} = 0, \quad Δ\mathcal{L} ≡ ∂_μ \left(iħ \bar{ψ} γ^μ Δψ\right)
\end{align}$$
where "≡" denotes "on-shell" equality, with $Δ\mathcal{L}$ reduced, on-shell, to the variational involving the involving the boundary term.
It should actually be made symmetric so that the boundary term gets contributions from $Δ\bar{ψ}$ as well as from $Δψ$:
$$\mathcal{L} = \bar{ψ} \left(iħ \overleftrightarrow{γ^μ ∂_μ} - mc\right) ψ,$$
where
$$\overleftrightarrow{γ^μ ∂_μ} = \frac{γ^μ \overrightarrow{∂_μ} - \overleftarrow{∂_μ} γ^μ}{2}.$$
Then
$$\begin{align}
\mathcal{L}
&= \frac{1}{2}\overbrace{\bar{ψ} \left(iħ γ^μ \overrightarrow{∂_μ} - mc\right) ψ}^{\overrightarrow{\mathcal{L}}} - \frac{1}{2}\overbrace{\bar{ψ} \left(iħ \overleftarrow{∂_μ} γ^μ + mc\right) ψ}^{\overleftarrow{\mathcal{L}}}\\
Δ\overrightarrow{\mathcal{L}}
&= Δ\left(\bar{ψ}\left(iħ γ^μ \overrightarrow{∂_μ} - mc\right) ψ\right)\\
&= Δ\bar{ψ} \left(iħ γ^μ \overrightarrow{∂_μ} - mc\right) ψ - \bar{ψ}\left(iħ \overleftarrow{∂_μ} γ^μ + mc\right) Δψ + ∂_μ \left(iħ \bar{ψ} γ^μ Δψ\right)\\
Δ\overleftarrow{\mathcal{L}}
&=Δ\left(\bar{ψ}\left(iħ \overleftarrow{∂_μ} γ^μ + mc\right) ψ\right)\\
&= \bar{ψ}\left(iħ \overleftarrow{∂_μ} γ^μ + mc\right) Δψ - Δ\bar{ψ} \left(iħ γ^μ \overrightarrow{∂_μ} - mc\right) ψ + ∂_μ \left(iħ Δ\bar{ψ} γ^μ ψ\right)\\
&⇒\\
Δ\mathcal{L}
&= Δ\bar{ψ} \left(iħ γ^μ \overrightarrow{∂_μ} - mc\right) ψ - \bar{ψ}\left(iħ \overleftarrow{∂_μ} γ^μ + mc\right) Δψ + ∂_μ \left(iħ \frac{\bar{ψ} γ^μ Δψ - Δ\bar{ψ} γ^μ ψ}{2}\right)
\end{align}.$$
Defining
$$\overleftrightarrow{γ^μ Δ} = \frac{γ^μ \overrightarrow{Δ} - \overleftarrow{Δ} γ^μ}{2},$$
the boundary term can be written:
$$∂_μ \left(\bar{ψ} iħ\overleftrightarrow{γ^μ Δ} ψ\right).$$
Then,
$$
Δ\mathcal{L}
= Δ\bar{ψ} \left(iħ γ^μ \overrightarrow{∂_μ} - mc\right) ψ - \bar{ψ}\left(iħ \overleftarrow{∂_μ} γ^μ + mc\right) Δψ + ∂_μ \left(\bar{ψ} iħ\overleftrightarrow{γ^μ Δ} ψ\right).
$$