smc

Since $ABCD$ is cyclic, opposite angles must sum to $180^{\circ }$. Therefore, $\angle ADC+\angle ABC=180^{\circ }$, and $\angle ABC=180^{\circ }-\angle ADC=180^{\circ }-68^{\circ }=112^{\circ }$. Notice also that $\angle ABC$ and $\angle CBE$ form a linear pair, and so they sum to $180^{\circ }$. Therefore, $\angle EBC=180^{\circ }-\angle ABC=180^{\circ }-112^{\circ }=68^{\circ },\boxed{68^{\circ }}$. Notice that the answer didn't even depend on $\angle BAD$.