Let $K$ be an algebraically closed field and $H$ a finite dimensional semisimple-cosemisimple Hopf $K$-algebra, and let $A$ be a finite dimensional $H$-module algebra and $H^{\ast}= Hom_{K}(H, K)$.
Is $A$ a direct summand of $A \# H \# H^\ast$ as $A$-bimodules?
This should be a comment, but I am not sure the OP reads comments, so...
Sorry, I still don't understand this. How is $\left(A\#H\right)\#H^{\ast}$ an $A$-bimodule? The only way $A$ can act on $\left(A\#H\right)\#H^{\ast}$ is by acting on the $A$ factor, from both sides. And it leads to a rather boring $A$-$A$-bimodule, canonically isomorphic to $\left(A\otimes H\right)\otimes H^{\ast}$ because multiplication doesn't matter. Of course this has $A$ as a direct summand, since $\left(A\otimes H\right)\otimes H^{\ast}\cong A\otimes \left(H\otimes H^{\ast}\right)$.
What I also fail to understand is how $\left(A\#H\right)\#H^{\ast}$ is defined in the first place. In order to define $B\#H$, I need $B$ to be a left $H$-module algebra. Now even if $A$ is a left $H$-module algebra, how is $A\#H$ a left $H^{\ast}$-module algebra?
$\text{A#H#H*}$ is isomorphic to $M_n(A)$. You may check Moss E.Sweedler's textbook: "Hopf algebras" to find the proof
