= Solution
For curvature $-1$, the <Poincare disc model> has metric and area element
$$
\boxed{ds^2=\frac{4(dx^2+dy^2)}{(1-x^2-y^2)^2}},\qquad dA=\frac{4r\,dr\,d\theta}{(1-r^2)^2}.
$$
The given minimizing-diameter property makes the distance from zero to Euclidean radius $R$ equal to
$$
\rho=\int_0^R\frac{2\,dr}{1-r^2}=\log\frac{1+R}{1-R},\qquad\boxed{R=\tanh(\rho/2)}.
$$
Radial symmetry therefore identifies the whole <hyperbolic circle in the Poincare disc>. Integrating its enclosed <hyperbolic area> gives
$$
A=\int_0^{2\pi}\int_0^R\frac{4r}{(1-r^2)^2}\,dr\,d\theta=4\pi\left(\frac1{1-R^2}-1\right)=\boxed{2\pi(\cosh\rho-1)}.
$$
For a geodesic <hyperbolic triangle> with angles $\alpha,\beta,\gamma$, the <Gauss-Bonnet theorem> states $A=\pi-(\alpha+\beta+\gamma)$, in particular $A\leq\pi$, with strict inequality for ordinary finite vertices. Let $d$ be the distance of the interior point $P$ to the nearest side. The open disc of radius $d$ about $P$ is contained in the triangle: a path from $P$ to an exterior point must first cross the boundary at distance at least $d$. A hyperbolic isometry carries its center to zero, so its area is the area already calculated. Consequently
$$
2\pi(\cosh d-1)\leq A\leq\pi,\qquad\boxed{d\leq\operatorname{arcosh}(3/2)}.
$$
This uses <hyperbolic area bounds an inscribed disc> and establishes the requested bound without assuming that $P$ is an incenter. A sharper universal triangle bound is possible, but is not needed here.
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