After a measurement, the system should be in the eigenstate of $\hat A$
with eigenvalue $a_i$ (assuming no degeneracy in the spectrum) since you know the outcome was $a_i$. Normally, the measurement does not preserve the norm so physically you know that your density matrix after the measurement should be
$$
\hat \rho' \propto \vert a_i\rangle\langle a_i\vert\, .
$$
where $\vert a_i\rangle$ is the eigenvector of $\hat A$ corresponding to the eigenvalue $a_i$.
Now, start with the projector $P$ and the density matrix $\rho$ written
explicitly as
\begin{align}
P&=\vert a_i\rangle\langle a_i\vert \, ,\\
\hat \rho &= \sum_j p_j \vert \psi_j\rangle \langle \psi_j\vert
\end{align}
and examine $P\hat \rho P$. Inserting the expressions yields explicitly
\begin{align}
\hat \rho'&= \sum_j p_j \vert a_i\rangle\langle a_i\vert \psi_j\rangle \langle \psi_j \vert a_i\rangle\langle a_i\vert \\
&= \vert a_i\rangle\langle a_i\vert \left(\sum_j p_j \vert \langle a_i\vert \psi_j\rangle\vert^2\right)\, ,\\
&=\vert a_i\rangle \langle a_i\vert \beta_i\, , \qquad \beta_i> 0
\end{align}
which is what you expect. Of course, as mentioned above, $\hat \rho'$ is no longer normalized - but that's to be expected because projectors are no norm-preserving operators.
In this example where there is no degeneracy, $\hat \rho'$ is a pure state. You can scratch your head as to whether or not this remains true if the eigenvalue $a_i$ occurs multiple times in the spectrum and, in particular, if this degeneracy changes the projector $P$.
