= Solution
Put $A=\mathbb H\setminus U$. The neighborhood assumptions at zero and infinity mean that $A$ is a bounded hull away from the starting boundary point. For $t<T$, the conjugacy map $h_t$ is the hydrodynamically normalized <mapping-out function> of the remaining forbidden hull after removing $K_t$. Its closure avoids the driver, so the <boundary derivative is an excursion avoidance probability> yields $0<\Sigma_t\leq1$. Thus $0<M_t\leq1$, and the <bounded local martingale criterion> makes the localized <martingales> uniformly bounded and uniformly integrable.
Choose localization times $\tau_n\uparrow T$, also bounded by $n$. Optional stopping gives $\mathbb E M_{\tau_n}=M_0$. The supplied terminal limit and continuity of the power function give
$$
M_{\tau_n}\longrightarrow\mathbf1_{\{T=\infty\}},\qquad
M_0=\Phi'(0)^{5/8}.
$$
Since these variables are bounded by one, dominated convergence yields
$$
\boxed{\mathbb P(\gamma_t\in U\text{ for all positive times})
=\mathbb P(T=\infty)=\Phi'(0)^{5/8}.}
$$
As usual for a chord starting at zero, the time-zero point is understood as the boundary <prime end> of $U$; the avoidance assertion concerns the subsequent curve. The boundedness step is what upgrades the <local martingale> calculation to an equality of probabilities.
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