So, using the measurement postulate for pure quantum states, you derived the formula:
$$\sum_i q_i \frac{P_j |\psi_i\rangle\langle\psi_i|P_j}{p_{j|i}} \tag{1}$$
where $p_{j|i} = \|P_j |\psi_i\rangle \|^2$ is the probability of measuring result $j$ starting from the state $|\psi_i\rangle$. And from this you would like to deduce the known formula:
$$\frac{P_j \rho P_j}{p_j} \tag{2}$$
where $p_j = \sum q_i p_{j|i}$ is the overall probability of measuring result $j$. But that doesn't work! If you check some examples, you will see that in general expressions $(1)$ and $(2)$ are not equal.
So $(1)$ is not correct, but why? Apparently you did everything right. You started from a statistical superposition of quantum states $|\psi_i\rangle$ with probabilities $q_i$. After the measurement the state $|\psi_i\rangle$ becomes $|\psi'_i\rangle = \frac{P_j |\psi_i\rangle}{\sqrt{p_{j|i}}}$. So we simply end up with the statistical superposition of the $|\psi'_i\rangle$, right?
Well, yes, but for the subtlety that the classical statistical probabilities have changed due to the measurement! By reading out the result $j$, we have gained information about our initial mix of quantum states and we need to update our weights $q_i \rightarrow q'_i$ to reflect this. For example, in the extreme case where $P_j |\psi_i\rangle = 0$ for some of the $i$, we know the system cannot have been in one of those quantum states to begin with, or we wouldn't have measured $j$ (and the fact that $(1)$ fails to be normalized in this case is an indication that something is not right indeed).
So, to understand better what is going on, and to find out what the correct $q'_i$ are, let us imagine that we repeat the experiment a large number $N$ of times, always preparing our initial mixed state in the same way: this is, after all, what probabilities are about. In $N_i = q_i N$ of the experiments, the initial quantum state is $|\psi_i\rangle$. Among those, the result $j$ will be measured in $N_{i,j} = q_i p_{j|i} N$ experiments after which the state will be in $|\psi'_i\rangle$. Now, we discard all experiments in which a result different from $j$ has been obtained. We are left with only $N'=\sum_i N_{i,j} = p_j N$ experiments, among which the portion of experiments that started in the state $|\psi_i\rangle$ and are now in the state $|\psi'_i\rangle$ is $\frac{N_{i,j}}{N'} = \frac{q_i p_{j|i}}{p_j}$. That's our $q'_i$!
Now, if in $(1)$ we replace $q_i$ by this $q'_i$ we do arrive at $(2)$.