Preliminaries
$Q$ function and normal ordering — If an operator $B$ (which may or may not be a state) can be put in normal order, write it as $B_N(a,a^\dagger)\equiv\sum_{ij} c_{ij} a^{\dagger i}a^j$, then its $Q$ function is simply
$$Q_B(\alpha)=\frac1\pi B_N(\alpha,\bar\alpha)=\frac1\pi\sum_{ij} c_{ij} \bar\alpha^i \alpha^j.$$
In other words, you get the expression for $Q$ by simply replacing $a\to\alpha$ and $a^\dagger\to\bar\alpha$.
This is easy to prove, as $\langle \alpha|B_N(a,a^\dagger)|\beta\rangle=B_N(\beta,\bar\alpha)$.
Note that here I said "$B$ can be put in normal order" and not "$B$ is normally ordered". This might seem like a subtlety but I think it helps to distinguish these concepts: $B$, as an operator, is neither normally nor antinormally ordered. The concepts of "ordering" only make sense when you refer to specific ways to write the operators, before interpreting $a,a^\dagger$ as actual operators acting on some space. See What does "not applying the CCR" mean exactly? for more details on this point.
$P$ function and antinormal ordering — Similarly if an operator $B$ can be put in antinormal order, write it as $B_{A}(a,a^\dagger)$, then $P_B(\alpha)=\frac1\pi B_{A}(\alpha,\bar\alpha)$.
Convolution relation between $P$ and $W$ — The standard way to introduce $P$ and $Q$ functions is explicitly via the expression $Q_\rho(\alpha) = \frac1\pi \langle\alpha|\rho|\alpha\rangle$, and implicitly via
$$\rho = \int \mathrm d^2\alpha P_\rho(\alpha) |\alpha\rangle\!\langle\alpha|.$$
These two are sufficient to get the expression with the convolution, as already shown in the other answer:
$$Q_\rho(\alpha) = \frac1\pi \langle\alpha|\rho|\alpha\rangle
= \frac1\pi\int \mathrm d^2\beta P_\rho(\beta) \lvert\langle\alpha|\beta\rangle\rvert^2
= \frac1\pi\int \mathrm d^2\beta P_\rho(\beta) e^{-|\alpha-\beta|^2}.$$
A similar way to relate $W$ and $Q$ via convolution with a Gaussian is also discussed in Show that the Husimi Q function equals $Q(\alpha)=\frac 1 \pi\langle \alpha |\hat \rho|\alpha\rangle$ from its relation with the Wigner function.
Some explicit examples might be useful to better understand this.
Example 1: $B=a^\dagger a$
Consider the operator $B\equiv a^\dagger a$. This is already written in normal order, $B_N(a,a^\dagger)=a^\dagger a$, and thus
$$Q_B(\alpha) = \frac1\pi |\alpha|^2.$$
Writing it in antinormal order, we instead get $B_{A}(a,a^\dagger)= aa^\dagger-1$, and thus
$$P_B(\alpha) = \frac1\pi (|\alpha|^2-1).$$
For consistency, we can check that we have the integral representation:
$$a^\dagger a = \int \mathrm d^2\alpha \left(\frac{|\alpha|^2-1}{\pi}\right) |\alpha\rangle\!\langle\alpha|,$$
which amounts to the identity
$$
n \delta_{n,m} =
\frac1\pi\int \mathrm d^2\alpha (|\alpha|^2-1) e^{-|\alpha|^2}
\frac{\alpha^n\bar\alpha^m}{\sqrt{n!m!}}.
$$
Proving this identity is easy enough switching to polar coordinates, and is essentially equivalent to what's used in this answer.
We can also check how $Q$ is correctly reproduced via convolution of $P$ with a Gaussian:
$$\frac1\pi \int \mathrm d^2\beta P_B(\beta) e^{-|\alpha-\beta|^2}
= \frac1{\pi^2}\int \mathrm d^2\beta (|\beta|^2-1)e^{-|\alpha-\beta|^2}
= \frac{1}{\pi}|\alpha|^2.$$
Example 2: $B=a$
Perhaps an even simple example is $B=a$. In this case normal and antinormal orderings coincide: $B_A(a,a^\dagger)=B_N(a,a^\dagger)=a$, and thus so do $P$ and $Q$ representations:
$$P_B(\alpha) = Q_B(\alpha) = \frac1\pi \alpha.$$
You can then also again check for consistency that the identities
$$a = \int \mathrm d^2\alpha \left(\frac{1}{\pi}\alpha\right)|\alpha\rangle\!\langle\alpha|, \\
\frac1\pi\alpha = \frac1\pi\int \mathrm d^2\beta \left(\frac1\pi \beta\right)e^{-|\alpha-\beta|^2}$$
are satisfied (the first one is the representation of $a$ via the $P$ function, and the second one is the representation of $Q$ via convolution of a Gaussian with $P$).
Example 3: $\rho=|0\rangle\!\langle 0|$
Let's consider an example that's less trivial: the vacuum state, $\rho=|0\rangle\!\langle 0|$.
In this case, it's less trivial to find normal/antinormal orderings for $\rho$, but they still exist.
For example, a tidy expression for the normal ordering, discussed in Relationship between normal-ordered vacuum state and parity operator, is
$|0\rangle\!\langle 0| = N(e^{-a^\dagger a})$, which in other words means
$$\rho_N(a,a^\dagger) = \sum_{k=0}^\infty \frac{(-1)^k}{k!} a^{\dagger k}a^k.$$
In fact, as shown here, we have the integral representations
$$
|0⟩\!⟨0| = \frac1\pi\int d^2\gamma e^{-|\gamma|^2/2}D(\gamma)
= \frac1\pi\int d^2\gamma e^{-\bar\gamma a} e^{\gamma a^\dagger}
= \frac1\pi\int d^2\gamma e^{-|\gamma|^2} e^{\gamma a^\dagger} e^{-\bar\gamma a},
$$
where $D(\gamma)\equiv e^{\gamma a^\dagger-\bar\gamma a}$ is the displacement operator, and I juggled with its standard identities to get the different expressions. Note these expressions are symmetrically, antinormally, and normally ordered, respectively, and thus immediately also provide us with the $P$ and $Q$ representations for the vacuum state:
$$
P_\rho(\alpha) = \frac1{\pi^2} \int \mathrm d^2\gamma e^{-\bar\gamma \alpha}e^{\gamma\bar\alpha} = \delta^2(\alpha),
$$
$$Q_\rho(\alpha)
= \frac1{\pi^2} \int \mathrm d^2\gamma e^{-|\gamma|^2} e^{\gamma\bar\alpha}e^{-\bar\gamma \alpha} = \frac1\pi e^{|\alpha|^2}.$$
Of course, these expressions coincide with the well-known ones that can be derived pretty much directly from the definitions of $Q$ and $P$ given at the beginning of the post.
