= Solution
Take the convention that the coordinates of <planar Brownian motion> are independent standard real <Brownian motions>, so its <infinitesimal generator> is $\tfrac12\Delta$. Define the forward <conformal Brownian clock>
$$
A(s)=\int_0^s|\phi'(B_r)|^2\,dr,\qquad
\widetilde T=A(T-),\qquad \tau=A^{-1}.
$$
Since a <conformal bijection> has nonzero derivative, $A$ is a strictly increasing continuous map from $[0,T)$ onto $[0,\widetilde T)$. The time used inside the original path is its inverse:
$$
\boxed{\tau:[0,\widetilde T)\longrightarrow[0,T),\qquad
\widetilde B_t=\phi(B_{\tau(t)}).}
$$
Thus the interval direction for $\tau$ is the inverse-clock direction.
Write $\phi=u+iv$. The <Cauchy-Riemann equations> and the <Itô formula> make $u(B_s),v(B_s)$ continuous <local martingales>, with
$$
d[u(B)]_s=d[v(B)]_s=|\phi'(B_s)|^2ds,
\qquad d[u(B),v(B)]_s=0.
$$
Indeed both components are <harmonic functions>, and their gradients are orthogonal with the same squared norm. After the <time change of a continuous process> by $\tau$, their <quadratic variations> are $t$ and their <quadratic covariation> is zero. The <Lévy characterization of multidimensional Brownian motion> therefore identifies $\widetilde B$ as <planar Brownian motion> started at $z'$, up to its lifetime.
It remains to identify that lifetime as the exit time, rather than merely produce a local Brownian path. Boundedness of $D$ gives $T<\infty$ <almost surely>. For every compact subset $C\subset D'$, its inverse image under $\phi$ is compactly contained in $D$. As $s\uparrow T$, continuity gives $B_s\to B_T\notin D$, so $\phi(B_s)$ eventually leaves $C$. Consequently the transformed path leaves every compact subset of $D'$ at its lifetime. If $\widetilde T<\infty$, its Brownian extension has a finite limit, and that limit is outside $D'$; if $\widetilde T=\infty$, the path never exits. In both cases its maximal lifetime is precisely the exit time from $D'$.
\b[The killed paths, together with their lifetimes, have the same law]:
$$
\boxed{\bigl(\widetilde T,(\widetilde B_t)_{t<\widetilde T}\bigr)
\overset{d}=\bigl(T',(B'_t)_{t<T'}\bigr).}
$$
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