In Quantum Mechanics how can I prove this property?
"Sandwiching" an operator between a bra and a ket is standard notation, but I'm going to use somewhat clearer notation and then introduce the "sandwiching" at the end to make this more clear.
The inner product between a state $\psi$ and a second state $\phi$ is written like this:
$$\langle \psi | \phi\rangle$$ If I operate on $\phi$ with the operator $A$ before I take the inner product, then I would write it like this:
$$\langle \psi | A\phi\rangle$$
On the other hand, if I operate on $\psi$ with the operator $B$ before I take the inner product, then I would write that like this:
$$\langle B \psi | \phi\rangle$$
Lastly, because of the properties of the inner product, taking the complex conjugate is equivalent to flipping the order of the states, so
$$ \langle \psi | \phi \rangle^* = \langle \phi | \psi \rangle$$
Now that that's been taken care of, the Hermitian conjugate of an operator $A$, which we denote by $A^\dagger$, is defined as follows: for any two states $\phi$ and $\psi$,
$$\langle\phi | A \psi\rangle = \langle A^\dagger \phi | \psi \rangle$$
Therefore, $$\langle \phi | A | \psi \rangle^* \equiv \langle \phi | A\psi\rangle^* = \langle A^\dagger \phi | \psi \rangle^* = \langle \psi | A^\dagger \phi \rangle \equiv \langle \psi | A^\dagger | \phi \rangle$$
where, at the beginning and the end, I've used the "sandwich" notation. Such notation is quite common and the reasons why it's questionable are fairly technical, so I wouldn't worry about them for now.