Why if an operator $A$ bounded then $A^{\ast}$ is bounded?
I have already proven that $\|A\|=\|A^{\ast}\|$ but if $A$ is bounded then exists $M$ such that $\|A(f)\|\leq M||f||$ for all $f$ in $H$ Hilbert. How prove $\|A^{\ast} f\|\leq M^{\ast} \|f\|$ for some $M^{\ast}$?
The adjoint has a more general meaning in the context of Banach Spaces, which I think is actually more intuitive than the Hilbert Space case.
Suppose $X$ and $Y$ are Banach spaces and $T : X \to Y$ is bounded and linear. Then, define $$T^* : Y^* \to X^* : f \mapsto f \circ T.$$ Then, because $T$ is bounded and $f$ is bounded, we have $$\|T^* f\| = \|f \circ T\| = \sup_{y \in B_Y} f(Ty) \le \sup_{y \in B_Y} \|f\|\|Ty\| \le \sup_{y \in B_Y} \|f\|\|T\|\|y\| = \|f\|\|T\| \tag{1},$$ implying that $\|T^*\|$ is bounded with $\|T^*\| \le \|T\|$.
In the case of a Hilbert space, we have that $X^* \sim X$ and $Y^* \sim Y$ by the Riesz representation theorem. In particular, the map from $x \in X$ to $\langle \cdot, x \rangle \in X^*$ is a surjective linear isometry; the operator norm of $\langle \cdot, x \rangle$ is precisely $\|x\|$.
So, when we compute $T^* y$, we are really computing $T^*(\langle \cdot, y \rangle)$, which is the bounded linear functional $$X \mapsto \Bbb{F} : v \mapsto \langle Tv, y \rangle.$$ Such a functional must have a unique $x \in X$ associated to it, and that $x$ is what we call $T^* y$. In particular, we must have $$\langle Tv, y \rangle = \langle v, x \rangle$$ for all $v$, which leads us to the defining identity for adjoints: $$\langle Tv, y \rangle = \langle v, T^* y \rangle.$$
So, in short, the reason why the adjoint is bounded is because of (1).