Is there a Banach space version of Riesz representation theorem?

In general, such a $T^{\ast}$ doesn't exist. For an example, let $Y$ a non-reflexive Banach space, $X$ its dual, and $B$ the evaluation map. Let $\eta \in Y^{\ast\ast}\setminus Y$ and $x_1 \in X \setminus \{0\}$. Define $T(x) = \eta(x)\cdot x_1$. Then the Banach space adjoint (or transpose) of $T$ is ${}^tT \colon X^{\ast} \to X^{\ast}$ given by ${}^tT(\xi) = \xi \circ T \colon x \mapsto \eta(x)\cdot \xi(x_1)$, i.e. ${}^tT(\xi) = \xi(x_1)\cdot \eta$. If $T^{\ast}$ existed, it would be the restriction of ${}^tT$ to $Y$, but for $y \notin \ker x_1$, we have ${}^tT(y) = x_1(y)\cdot \eta \notin Y$.

Generally, as vector spaces, we can identify $Y$ with a subspace of $X^{\ast}$, and have $B$ the restriction of the evaluation map to $X\times Y$. Then the existence of $T^{\ast} \colon Y \to Y$ satisfying $B(Tx,y) = B(x,T^{\ast}y)$ for all $x,y$ means that ${}^tT(Y) \subset Y$, where ${}^tT$ is the transpose of $T$. That is the case if and only if $T$ is continuous with respect to $\sigma(X,Y)$. In that case, the closed graph theorem yields the continuity of $T^{\ast} = {}^tT\lvert_Y$ with respect to the chosen norm on $Y$ (which generally isn't equivalent to the norm induced by $X^{\ast}$).