Trace of density matrix for mixed state $\DeclareMathOperator{\Tr}{Tr}$On page 5 of this online document, it states a seemingly trivial fact: that if we have a density-matrix for a mixed state defined by
$$\hat{\rho}=\sum_kp_k|\psi_k\rangle\langle\psi_k|$$
where $\{|\psi_k\rangle\}$ are (not-necessarily orthogonal) pure states, then we have the following double-sided implication:
$$\Tr (\hat{\rho})=1~~~\iff~~~\sum_kp_k=1$$
This seems intuitively clear to me, but when I try to go from the left-side to the right-side I get stuck. Here's what I mean:
$$\begin{align}
\Tr(\hat{\rho})&=\sum_m \langle\psi_m| \hat{\rho }| \psi_m \rangle \\
&=\sum_{m} \langle\psi_m|\left(\sum_k p_k|\psi_k\rangle\langle\psi_k|\right)| \psi_m \rangle\\
&=\sum_k p_k \sum_m |\langle \psi_m |\psi_k\rangle |^2
\end{align}$$
Now, if $\{| \psi_k\rangle \}$ is orthogonal, then $|\langle \psi_m |\psi_k\rangle |^2=\delta_{mk}$ and everything works out easily - but, they aren't orthogonal. So what do I do?
 A: Let us focus on your chain of identities.
$$\begin{align}
\text{Tr}(\hat{\rho})&=\sum_m \langle\psi_m| \hat{\rho }| \psi_m \rangle \\
&=\sum_{m} \langle\psi_m|\left(\sum_k p_k|\psi_k\rangle\langle\psi_k|\right)| \psi_m \rangle\\
&=\sum_k p_k \sum_m |\langle \psi_m |\psi_k\rangle |^2
\end{align}$$
The point in the implications above is that the first line is the correct definition of trace if and only if the vectors $| \psi_m \rangle$ form a orthonormal basis. Otherwise the right-hand side is not the trace of $\hat{\rho}$ and the reasoning stops there. 
If the vectors $| \psi_m \rangle$ are normalized but are not mutually orthogonal and $$\hat{\rho} :=\sum_k p_k|\psi_k\rangle\langle\psi_k|\:,$$
then the correct procedure is to pick out an orthonormal basis of vectors $| \phi_m \rangle$ and then
$$\begin{align}
\text{Tr}(\hat{\rho})&=\sum_m \langle\phi_m| \hat{\rho }| \phi_m \rangle \\
&=\sum_{m} \langle\phi_m|\left(\sum_k p_k|\psi_k\rangle\langle\psi_k|\right)| \phi_m \rangle\\
&=\sum_k p_k \sum_m |\langle \phi_m |\psi_k\rangle |^2 = \sum_k p_k |||\psi_k\rangle||^2 = \sum_k p_k 1 = \sum_k p_k
\end{align}$$
we therefore have the wanted  double-sided implication you mention:
$$ \text{Tr} (\hat{\rho})=1~~~\iff~~~\sum_kp_k=1\:.$$
