The P-function of a state $\rho$ (focusing on the single-mode case) can be written as, using a notation analogous to the one in Gerry&Knight' book, $$P_\rho(\alpha) = \int \frac{d^2\eta}{\pi^2} \chi_N(\eta) e^{\alpha\bar\eta-\bar\alpha\eta}, \quad \chi_N(\eta)\equiv \operatorname{tr}[\rho\exp(\eta a^\dagger) \exp(-\bar\eta a)].$$ This also means that we should be able to write $P_\rho(\alpha)$ as the expectation value of some observable on $\rho$, i.e. $P_\rho(\alpha) = \operatorname{tr}(\mathcal O_P(\alpha)\rho)$, where I defined $$\mathcal O_P(\alpha) \equiv \int \frac{d^2\eta}{\pi^2} \exp(\eta A_\alpha^\dagger)\exp(-\bar\eta A_\alpha), \quad A_\alpha\equiv a-\alpha.$$ For reference, the same procedure is especially straightforward with the $Q$ function instead, and leads to $$\mathcal O_Q(\alpha) \equiv \int \frac{d^2\eta}{\pi^2} e^{-\bar\eta A_\alpha}e^{\eta A_\alpha^\dagger} = \int\frac{d^2\beta}{\pi}\int\frac{d^2\eta}{\pi^2} \mathbb{P}_\beta e^{-\bar\eta(\beta-\alpha)+\eta(\beta-\alpha)^*} = \frac1\pi \mathbb{P}_\alpha,$$ where I integrated over $\eta$ in the last step, and used the shorthand notation $\mathbb{P}_\beta\equiv|\beta\rangle\!\langle\beta|$. Of course, this recovers the well-known fact that $Q_\rho(\alpha)=\frac{\langle\alpha|\rho|\alpha\rangle}{\pi}$.
What I'm trying to figure out is whether there's any kind of similar calculation/simplification that can be done for $\mathcal O_P$. We can't simply add a $\mathbb{P}_\beta$ in between the operator exponentials, as they're in the wrong order. We can add two of them on left and right, but that would lead to $$\mathcal O_P(\alpha) = \int\frac{d^2\eta}{\pi^2} \int\frac{d^2\beta d^2\gamma}{\pi^2} \mathbb{P}_\beta\mathbb{P}_\gamma e^{\eta(\beta-\alpha)^*-\bar\eta(\gamma-\alpha)}.$$ The problem is that now the exponential over $\eta$ is problematic, and generally leads to exponential divergences unless $\beta,\gamma$ are suitably chosen. An alternative approach would be to exploit the usual commutation rules to write $$\mathcal O_P(\alpha) = \int \frac{d^2\eta}{\pi^2} e^{-\bar\eta A_\alpha}e^{\eta A_\alpha^\dagger} e^{|\eta|^2} = \int\frac{d^2\beta}{\pi}\mathbb{P}_\beta \int\frac{d^2\eta}{\pi^2} e^{\eta(\beta-\alpha)^*-\bar\eta(\beta-\alpha)+|\eta|^2}. $$ This looks slightly more "elegant", but throws the divergence problem in my face even stronger with the $\exp(|\eta|^2)$ term. Decomposed in real and imaginary parts, the argument in the exponential reads $$2i[\eta_2(\beta_1-\alpha_1)-\eta_1(\beta_2-\alpha_2)] + \eta_1^2+\eta_2^2,$$ which seems like it could never converge. Though I'm guessing convergence is still possible thanks to the exponential terms hidden $\mathbb{P}_\beta$ which might kill the $\eta$ terms. After all, we know that, for example, $\operatorname{tr}(\mathbb{P}_\gamma \mathcal O_Q(\alpha))=\delta^2(\alpha-\gamma)$.
Now, I'm aware that the $P$ function is highly irregular (can be more singular than a $\delta$ etc), so it is absolutely to be expected that $\mathcal O_P$ would be precisely as irregular as $P$. Still, singularities can often be dealt with and written concisely, at least at a formal level. Is there any way to simplify these expressions, or maybe put them in a form that makes it more evident that the singularities disappear in at least some cases?
