= Solution
We prove necessity of the geometric condition from the embedding condition. Use normalized angular measure $m$ and the <H1 space on the unit disk> norm $\|f\|_1=\sup_{r<1}\int|f(r\xi)|\,dm(\xi)$. Testing $f=1$ first gives $\mu(\mathbb D)\le N(\mu)$.
For a short arc $I$ of normalized length $\ell=m(I)$ centered at $\xi_0$, put $a=(1-\ell)\xi_0$ and use the <normalized squared Cauchy kernel>
$$
f_a(z)=\frac{1-|a|^2}{(1-\overline a z)^2}.
$$
Its boundary modulus is the <Poisson kernel on the circle>, so $\|f_a\|_1=1$. The <geodesic-cap and Carleson-box comparison> gives $|z-\xi_0|\le C\ell$ for $z\in Q(I)$. Therefore $|1-\overline a z|\le C'\ell$, whereas $1-|a|^2=2\ell-\ell^2\ge\ell$. Thus $|f_a(z)|\ge c/\ell$ throughout the <hyperbolic geodesic cap>. The embedding bound gives
$$
\frac c\ell\mu(Q(I))\le\int_{\mathbb D}|f_a|\,d\mu\le N(\mu).
$$
For arcs longer than a fixed small threshold, the total-mass estimate gives the same form after changing the absolute constant. Hence
$$
\boxed{\mu(Q(I))\le C N(\mu)m(I).}
$$
This is precisely the <Carleson measure> condition, in its geodesic-cap version.
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