A common standard model lagrangian is written in a cup like this. It appears in many places also on a T shirt. But isnt that there is an obvious mistake?
That the Dirac lagrangian is already itself hermitian conjugation.
Thus we only need $$i \bar{\psi} \displaystyle{\not}D \psi$$ instead of $$i \bar{\psi} \displaystyle{\not}D \psi +h.c.$$ on the pictures?
It is a mistake. It's true, as FrodCube said, that you can work around it by redefining the field, but no one actually defines the field that way.
The origin of it is a CERN photo session of John Ellis "working on the blackboard"; obviously he is not really working but just drew some pretty pictures to pose in front of.
CERN sells various souvenirs containing the formula in Ellis's handwriting. You can see that the mistake has been fixed in some but not others. Confusingly, it seems to have been fixed in the thumbnail image of the mug, but not in the larger image on the product page. Fixed versions of the CERN mug definitely exist in the wild: see this tweet for instance. I suspect the T-shirt in your question is a third-party product that copied CERN's formula.
A Quantum Diaries post pointed out the mistake a decade ago. In the comments, someone said that according to a CERN physicist, the extra h.c. stands for "hot coffee".
It's not really a mistake, just a non-canonical normalization for $\psi$. The physics with and without the $h.c.$ is literally the same.
The canonical way of normalizing $\psi$ is such that
$$\mathcal L = i \bar\psi \not\partial \psi$$
Adding the $h.c.$ as in the t-shirt, since this term is Hermitean is equivalent to having $$\mathcal L = 2 i \bar\psi \not\partial \psi$$
This is just a different normalization for $\psi$ that can be elminated by redefining
$$\psi \to \psi/\sqrt2$$
Still, it's a "mistake" in the sense that it's a typo.
Also that Lagrangian is written in an extremely simplified way, the $h.c.$ is the smallest problem.
What was probably intended by the "h.c." with the fermion kinetic term is what should actually be written like this: $$\overline{ψ} \frac{γ^μ \left(+iħ \overrightarrow{∂_μ} + A^a_μ T_a\right) + \left(-iħ \overleftarrow{∂_μ} + T_a A^a_μ\right)γ^μ}2 ψ. $$ for suitably-defined representations $T_a$ of the gauge group basis. Normally, you'd write it like this: $$\overline{ψ} γ^μ \left(iħ \overleftrightarrow{∂_μ} + A^a_μ T_a\right) ψ,\quad \overleftrightarrow{∂_μ} = \frac{\overrightarrow{∂_μ} - \overleftarrow{∂_μ}}2 $$ but you want to get the $γ^μ$ out of the way of the left and right derivative operators, if you're in non-Cartesian/Minkowski curvilinear coordinates. Specifically: the "+ h.c." part was probably meant to convey that the derivative operator is to act evenly on the left and right.
