= Solution
Put $a=1/4$ and $E-a=\{\gamma-a:\gamma\in E\}$. Both events consist of loops surrounding $0$; $E-a$ lies in the disc of radius $3/4$, so its loops remain in $D$. Translation invariance and the pointed representation imply
$$
C\,\rho\{\Gamma(B)\in E\}=\mu(E)=\mu(E-a)
=C\,\rho\{\Gamma(B)\in E-a\}.
$$
Since $C>0$, cancel it. Outer boundaries commute with translation, and $\Gamma(B)\in E-a$ is exactly $\Gamma(B+a)\in E$. Hence
$$
\boxed{\rho\{\Gamma(B)\in E\}=\rho\{\Gamma(B+1/4)\in E\}.}
$$
The half-disc condition ensures that translating a relevant filled Brownian trace still lies inside the original unit-disc cutoff. This proves the asserted event identity; it does not claim translation invariance of the entire measure rooted at $0$.
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