Step 1: Recall what A, B, C, D actually mean.
\[ V_1=AV_2-BI_2,\qquad I_1=CV_2-DI_2 \]
$A$ and $C$ are found with port 2 open ($I_2=0$), while $B$ and $D$ are found with port 2 shorted ($V_2=0$).
Step 2: Open port 2 and find $A$ and $C$.
With $I_2=0$, no current flows in the last $1k\Omega$ series arm, so it drops no voltage and $V_2$ equals the voltage across the middle $1k\Omega$ shunt arm. A current $I_1$ entering port 1 flows in through the first series resistor and then entirely down the shunt arm (there is nowhere else for it to go). So the shunt arm carries $I_1$ and
\[ V_2=I_1(1000) \]
Also $V_1=I_1(1000)+V_2=I_1(1000)+I_1(1000)=2000I_1$. Then
\[ A=\frac{V_1}{V_2}\bigg|_{I_2=0}=\frac{2000I_1}{1000I_1}=2 \]
\[ C=\frac{I_1}{V_2}\bigg|_{I_2=0}=\frac{I_1}{1000I_1}=\frac{1}{1000}=10^{-3}\ \Omega^{-1} \]
Step 3: Short port 2 and find $B$ and $D$.
With $V_2=0$, the last $1k\Omega$ arm is now directly in parallel with the middle $1k\Omega$ shunt arm, since both ends of the output arm sit at the same potential as the bottom rail once shorted. That parallel combination is $1000\parallel1000=500\Omega$, in series with the first $1k\Omega$ arm, so a source $V_1$ sees a total of $1500\Omega$ and drives $I_1=V_1/1500$. The short-circuit output current $I_2$ splits evenly between the shunt arm and the shorted series arm, because both are $1000\Omega$: $I_2=I_1/2$.
So
\[ B=\frac{V_1}{I_2}\bigg|_{V_2=0}=\frac{1500I_1}{I_1/2}=3000\ \Omega=3\times10^3\ \Omega \]
\[ D=\frac{I_1}{I_2}\bigg|_{V_2=0}=\frac{I_1}{I_1/2}=2 \]
