Remark: I assume that you want $A$ to be a non-trivial bundle. Otherwise, of course, any parallelizable compact manifold would be an example. In particular, any compact Lie group would be an example with a nice compact group action.

A simple interesting example is $S^5$, whose tangent bundle can be written as the sum of a line bundle and an irreducible 4-plane bundle. It also has a nice compact group action, as $\mathrm{SO}(6)$ acts transitively on it in the obvious way.

The proof of this is as follows: First, $S^5$ is an odd dimensional sphere, so it has a nonvanishing tangent vector field. Hence $TS^5 = L_1 \oplus A$ where $L_1$ is a trivial line bundle and $A$ is a 4-plane bundle. It remains to show that $A$ is irreducible. First, $A$ cannot contain any non-trivial line bundle $L_2$ since $L_2$ would have to be orientable (since $S^5$ is simply connected) and hence trivial, and a non-vanishing section of such an $L_2$ would yield a tangent vector field on $S^5$ that is linearly independent from any non-vanishing section of $L_1$. However, as is well-known, $S^5$ cannot have two linearly independent tangent vector fields. Second, $A$ cannot split as the sum of two $2$-plane bundles since $H^2(S^5,Z)$ is trivial, so any $2$-plane bundle over $S^5$ must be trivial and hence a sum of two (trivial) line bundles. Hence, if $A$ could be written as the sum of two $2$-plane bundles, $S^5$ would have to be parallelizable, which, we know, it is not. Thus, $A$ is irreducible.

Finally, note that $S^5$ admits a (transitive) compact group action by $\mathrm{O}(6)$, which is as nice as possible.

Remark: I assume that you want $A$ to be a non-trivial bundle. Otherwise, of course, any parallelizable compact manifold would be an example. In particular, any compact Lie group would be an example with a nice compact group action.

A simple interesting example is $S^5$, whose tangent bundle can be written as the sum of a line bundle and an irreducible 4-plane bundle. It also has a nice compact group action, as $\mathrm{SO}(6)$ acts transitively on it in the obvious way.

The proof of this is as follows: First, $S^5$ is an odd dimensional sphere, so it has a nonvanishing tangent vector field. Hence $TS^5 = L_1 \oplus A$ where $L_1$ is a trivial line bundle and $A$ is a 4-plane bundle. It remains to show that $A$ is irreducible. First, $A$ cannot contain any non-trivial line bundle $L_2$ since $L_2$ would have to be orientable (since $S^5$ is simply connected) and hence trivial, and a non-vanishing section of such an $L_2$ would yield a tangent vector field on $S^5$ that is linearly independent from any non-vanishing section of $L_1$. However, as is well-known, $S^5$ cannot have two linearly independent tangent vector fields. Second, $A$ cannot split as the sum of two $2$-plane bundles since $H^2(S^5,Z)$ is trivial, so any $2$-plane bundle over $S^5$ must be trivial and hence a sum of two (trivial) line bundles. Hence, if $A$ could be written as the sum of two $2$-plane bundles, $S^5$ would have to be parallelizable, which, we know, it is not. Thus, $A$ is irreducible.