= Solution
We prove sufficiency by a <non-tangential maximal function> and the <layer cake representation>. Write $K=C(\mu)$. First $\mu(\mathbb D)\le C K$: three <hyperbolic geodesic caps> corresponding to arcs of normalized length $2/3$, with equally spaced centers, cover the <unit disc>. This handles interior points and large arcs, not only a thin boundary strip.
Fix aperture two and set $Nf(\xi)=\sup_{z\in\Gamma_2(\xi)}|f(z)|$. The crucial <H1 non-tangential maximal inequality on the disk> is
$$
\int_{\partial\mathbb D}Nf\,dm\le C\|f\|_1.
$$
Here is its endpoint proof. For $0<R<1$ let $f_R(z)=f(Rz)$, which is <holomorphic> across the closed <unit disc>. Since <modulus powers of holomorphic functions are subharmonic>, Poisson comparison gives
$$
|f_R(z)|^{1/2}\le P[g_R](z),\qquad g_R(\xi)=|f(R\xi)|^{1/2}.
$$
The estimate that a <Poisson maximal function is dominated by the Hardy-Littlewood maximal function> yields $(Nf_R)^{1/2}\le C M g_R$. The strong $L^2$ <Hardy-Littlewood maximal inequality> now gives
$$
\int Nf_R\,dm\le C\int(Mg_R)^2\,dm
\le C\int|f(R\xi)|\,dm\le C\|f\|_1.
$$
For each interior point $f_R(z)\to f(z)$, hence $Nf\le\liminf_{R\uparrow1}Nf_R$. The <Fatou lemma> proves the maximal inequality. This uses a genuine $L^2$ maximal bound; the <Hardy-Littlewood maximal function> is not strongly bounded on $L^1$.
For $t>0$, the set $O_t=\{\xi:Nf(\xi)>t\}$ is open, since each interior point witnessing this strict inequality has an open boundary shadow. Write its disjoint arc components as $I_j$. If $|f(z)|>t$ and $\delta=1-|z|$, the arc centered at $z/|z|$ with angular half-width $\delta$ lies in $O_t$, because $|z-\xi|<2\delta$ there. Thus it lies in one component $I_j$, and $z$ has angular projection in $I_j$ with $\delta\le\pi m(I_j)$. The <geodesic-cap and Carleson-box comparison> puts all such points in $Q(\kappa I_j)$ for a fixed enlargement factor $\kappa$ whenever $I_j$ is small. For large $I_j$, use $\mu(\mathbb D)\le C K\le C'K m(I_j)$. If $z=0$ or $O_t$ is the entire circle, that same total-mass estimate applies. Summing, without needing the enlarged caps to be disjoint, gives
$$
\mu\{z:|f(z)|>t\}\le C K\sum_jm(I_j)=C K m(O_t).
$$
The <layer cake representation> and the proved <H1 non-tangential maximal inequality on the disk> finish the argument:
$$
\int_{\mathbb D}|f|\,d\mu
=\int_0^\infty\mu\{|f|>t\}\,dt
\le C K\int_0^\infty m\{Nf>t\}\,dt
\le C K\|f\|_1.
$$
Thus \b[the two conditions are equivalent], with their least admissible constants comparable: $C(\mu)\le C_1N(\mu)$ and $N(\mu)\le C_2C(\mu)$. This is the <Carleson embedding theorem on the disk>, proved at the analytic $H^1$ endpoint.
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