As nothing is mentioned in question, assuming that the first state itself is the start state i.e. State $A$ so, State $G$ is not reachable from the start and hence is removed before applying the minimization algorithm.
- $P0 \rightarrow [ABEFG] [CDH] \rightarrow [ABEF] [CDH]$
- $P1 \rightarrow [AB] [EF] [CDH]$
- $P2 \rightarrow [A] [B] [EF] [CD] [H]$
- $P3 \rightarrow [A] [B] [EF] [CD] [H]$
//$P2==P3$ So, stop
State table for the minimal machine ::
$$\begin{array}{|l|l|l|} \hline \text{Present State} & \text{input = 0} & \text{input = 1} \\\hline \text{A} & \text{B.1} & \text{H.1} \\\hline \text{B} & \text{EF.1} & \text{CD.1} \\\hline \text{CD} & \text{CD.0} & \text{EF.1} \\\hline \text{EF} & \text{CD.1} & \text{CD.1} \\\hline \text{H} & \text{CD.0} & \text{A.1} \\\hline \end{array}$$