= Solution
For $r>0$, let $\kappa_r$ be the <probability> law of a seed displacement from its tree. For the literal circular support, it is normalized <arclength> on $\{u:|u|=r\}$. The conditional <mean measure of a point process> is
$$
\boxed{\Lambda_\Pi(A)=\mu\sum_{z\in\Pi}\kappa_r(A-z).}
$$
We prove the conditional Poisson assertion directly by joint <probability generating functions>.
Fix disjoint bounded <measurable> sets $A_1,\ldots,A_k$. For a tree at $z$, put $q_j(z)=\kappa_r(A_j-z)$ and $q_0(z)=1-\sum_jq_j(z)$. Conditional on its seed count, the allocation among these sets and their complement is <multinomial distribution>. Averaging over the Poisson seed count gives
$$
\mathbb E\left[\prod_{j=1}^k s_j^{N_z(A_j)}\,\middle|\,\Pi\right]
=\exp\left[\mu\left(q_0(z)+\sum_jq_j(z)s_j-1\right)\right]
=\prod_{j=1}^k\exp\bigl(\mu q_j(z)(s_j-1)\bigr).
$$
Thus this tree contributes <independent> Poisson counts of means $\mu q_j(z)$ to the disjoint sets. Different trees contribute independently. Only trees within distance $r$ of the bounded union of the $A_j$ can contribute, and there are <almost surely> finitely many of them. Multiplying their <probability generating functions> gives
$$
\mathbb E\left[\prod_j s_j^{\Pi^*(A_j)}\,\middle|\,\Pi\right]
=\prod_j\exp\bigl(\Lambda_\Pi(A_j)(s_j-1)\bigr).
$$
This is precisely the joint <probability generating function> of <independent> Poisson counts with the displayed means. Also $\Lambda_\Pi$ is locally finite, because each bounded set can receive seeds from only finitely many trees. This proves the conditional <Poisson point process> assertion and the <Poisson offspring clusters directed by their parent process> construction.
In <arclength> notation the directing measure is $\Lambda_\Pi=\mu\sum_{z\in\Pi}\sigma_{z,r}/(2\pi r)$, where $\sigma_{z,r}$ is <arclength> on the circle centered at $z$. If “circle” is used to mean the filled disc instead, take $\kappa_r(du)=\mathbf1_{\{|u|\le r\}}du/(\pi r^2)$; then
$$
\Lambda_\Pi(dx)=\frac{\mu}{\pi r^2}\sum_{z\in\Pi}\mathbf1_{\{|x-z|\le r\}}\,dx.
$$
The generating-function proof and the following unconditional conclusion hold for either reading. The circular version has a singular conditional directing measure, giving the <random directing measure of a Cox process> formulation.
Unconditionally this is a <Cox process>, specifically a Poisson-offspring <Neyman-Scott process>. Its <mean measure of a point process> is nevertheless spatially uniform. With $q_A(z)=\kappa_r(A-z)$, the <Campbell first-moment formula> and translation invariance give
$$
\mathbb E\Lambda_\Pi(A)=\lambda\mu\int_{\mathbb R^2}q_A(z)\,dz
=\lambda\mu|A|,
$$
since $\int q_A(z)dz=\int\kappa_r(du)\int\mathbf1_A(z+u)dz=|A|$. Uniform mean alone does not establish a Poisson law.
For a bounded set $A$ of positive area, the conditional count has mean and <variance> $\Lambda_\Pi(A)$. The <law of total variance> gives
$$
\operatorname{Var}\Pi^*(A)=\mathbb E\Lambda_\Pi(A)+\operatorname{Var}\Lambda_\Pi(A).
$$
A Poisson integral of a deterministic function $q$ has <variance> $\lambda\int q^2dz$: for a simple function on disjoint sets this follows from <independent> Poisson counts, and square-integrable approximation gives the general formula. Here $0\le q_A\le1$ and it has bounded support, so
$$
\boxed{\operatorname{Var}\Pi^*(A)
=\lambda\mu|A|+\lambda\mu^2\int_{\mathbb R^2}q_A(z)^2\,dz
>\lambda\mu|A|=\mathbb E\Pi^*(A).}
$$
The strict inequality holds when $\lambda,\mu>0$, because $\int q_A=|A|>0$ implies $\int q_A^2>0$. A Poisson <random variable> has equal mean and <variance>. Hence \b[the unconditional process is not a <Poisson process> in the nontrivial model]. If $\lambda=0$ or $\mu=0$, there are no seeds and the process is the degenerate empty <Poisson process>. The <overdispersion of nondegenerate mixed Poisson counts> quantifies the additional clustering induced by the random parents.
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