We construct a locally risk-free self-financing portfolio $X_t$, at time $t$, with $\Delta_t^1$ share of debt and $\Delta_t^2$ share of equity. That is,
\begin{align*}
X_t = \Delta_t^1 D_t + \Delta_t^2 E_t.
\end{align*}
Then,
\begin{align*}
dX_t &=\Delta_t^1 dD_t + \Delta_t^2 dE_t\\
&=\Delta_t^1\bigg[\Big(\frac{\partial D_t}{\partial t} + \mu A_t\frac{\partial D_t}{\partial A_t} + \frac{1}{2} \sigma^2 A_t^2 \frac{\partial^2 D_t }{\partial A_t^2}\Big)dt + \sigma A_t \frac{\partial D_t}{\partial A_t} dW_t \bigg]\\
&\quad + \Delta_t^2\bigg[\Big(\frac{\partial E_t}{\partial t} + \mu A_t \frac{\partial E_t}{\partial A_t} + \frac{1}{2} \sigma^2 A_t^2 \frac{\partial^2 E_t }{\partial A_t^2}\Big)dt + \sigma A_t \frac{\partial E_t}{\partial A_t} dW_t \bigg].
\end{align*}
Since $X_t$ is locally risk-free,
\begin{align*}
\Delta_t^1\frac{\partial D_t}{\partial A_t} + \Delta_t^2\frac{\partial E_t}{\partial A_t}=0. \tag{1}
\end{align*}
Moreover, since $X_t$ earn the risk-free rate $r$,
\begin{align*}
dX_t = rX_t dt.
\end{align*}
From $(1)$,
\begin{align*}
r\Delta_t^1 D_t dt + r\Delta_t^2 E_t dt &= \Delta_t^1\Big(\frac{\partial D_t}{\partial t} + \mu A_t \frac{\partial D_t}{\partial A_t} + \frac{1}{2} \sigma^2 A_t^2 \frac{\partial^2 D_t }{\partial A_t^2}\Big)dt \\
&\quad + \Delta_t^2\Big(\frac{\partial E_t}{\partial t} + \mu A_t \frac{\partial E_t}{\partial A_t} + \frac{1}{2} \sigma^2 A_t^2 \frac{\partial^2 E_t }{\partial A_t^2}\Big)dt.
\end{align*}
That is,
\begin{align*}
&\Delta_t^1\Big(\frac{\partial D_t}{\partial t} + \mu A_t \frac{\partial D_t}{\partial A_t}+\frac{1}{2} \sigma^2 A_t^2 \frac{\partial^2 D_t }{\partial A_t^2} - rD_t \Big)dt \\
&+ \Delta_t^2\Big(\frac{\partial E_t}{\partial t} +\mu A_t \frac{\partial E_t}{\partial A_t}+ \frac{1}{2} \sigma^2 A_t^2 \frac{\partial^2 E_t }{\partial A_t^2}-rE_t\Big)dt =0. \tag{2}
\end{align*}
Since $A_t = E_t + D_t$,
\begin{align*}
\frac{\partial D_t}{\partial t} &= -\frac{\partial E_t}{\partial t},\tag{3}\\
\frac{\partial D_t}{\partial A_t} &= 1- \frac{\partial E_t}{\partial A_t},\tag{4}\\
\frac{\partial^2 D_t }{\partial A_t^2} &= - \frac{\partial^2 E_t }{\partial A_t^2}.\tag{5}
\end{align*}
From $(1)$ and $(2)$,
\begin{align*}
&-\Delta_t^2\frac{\frac{\partial E_t}{\partial A_t}}{\frac{\partial D_t}{\partial A_t}}\Big(\frac{\partial D_t}{\partial t} + \mu A_t \frac{\partial D_t}{\partial A_t}+\frac{1}{2} \sigma^2 A_t^2 \frac{\partial^2 D_t }{\partial A_t^2} - rD_t \Big) \\
& \quad + \Delta_t^2\Big(\frac{\partial E_t}{\partial t} +\mu A_t \frac{\partial E_t}{\partial A_t}+ \frac{1}{2} \sigma^2 A_t^2 \frac{\partial^2 E_t }{\partial A_t^2}-rE_t\Big) =0.
\end{align*}
Then, from $(3)$-$(5)$, and multiplying $\frac{\partial D_t}{\partial A_t}=1-\frac{\partial E_t}{\partial A_t}$ on both sides,
\begin{align*}
&-\frac{\partial E_t}{\partial A_t}\bigg[-\frac{\partial E_t}{\partial t} + \mu A_t \Big(1-\frac{\partial E_t}{\partial A_t}\Big)-\frac{1}{2} \sigma^2 A_t^2 \frac{\partial^2 E_t }{\partial A_t^2} - r(A_t-E_t) \bigg] \\
& \quad + \Big(1- \frac{\partial E_t}{\partial A_t}\Big)\Big(\frac{\partial E_t}{\partial t} +\mu A_t \frac{\partial E_t}{\partial A_t}+ \sigma^2 \frac{1}{2} A_t^2 \frac{\partial^2 E_t }{\partial A_t^2}-rE_t\Big) \\
=&\ \frac{\partial E_t}{\partial A_t}\frac{\partial E_t}{\partial t} - \mu A_t \frac{\partial E_t}{\partial A_t} + \mu A_t \left(\frac{\partial E_t}{\partial A_t}\right)^2 + \frac{1}{2} \sigma^2 A_t^2\frac{\partial E_t}{\partial A_t}\frac{\partial^2 E_t }{\partial A_t^2} +r(A_t-E_t)\frac{\partial E_t}{\partial A_t} \\
&\quad - \frac{\partial E_t}{\partial A_t}\frac{\partial E_t}{\partial t} -\mu A_t \left(\frac{\partial E_t}{\partial A_t}\right)^2 - \frac{1}{2} \sigma^2 A_t^2\frac{\partial E_t}{\partial A_t}\frac{\partial^2 E_t }{\partial A_t^2}+rE_t\frac{\partial E_t}{\partial A_t} \\
&\quad + \frac{\partial E_t}{\partial t} +\mu A_t \frac{\partial E_t}{\partial A_t}+ \frac{1}{2} \sigma^2 A_t^2 \frac{\partial^2 E_t }{\partial A_t^2}-rE_t \\
=& \ \frac{\partial E_t}{\partial t} +r A_t \frac{\partial E_t}{\partial A_t}+ \frac{1}{2} \sigma^2 A_t^2 \frac{\partial^2 E_t }{\partial A_t^2}-rE_t =0.\tag{6}
\end{align*}
Since $A_t$ also satisfies $(6)$, then so does $D_t$. The derivation is this complete.
