Proof of strong convexity of trace distance

I'm trying to follow the Nielsen and Chuang proof (equation 9.49 of Chapter 9, page 408). I reproduce it here for completeness.

With trace distance defined as $$D(\rho, \sigma) = \frac{1}{2}tr(|\rho - \sigma|)$$, it is known that there exists a projector $$P$$, such that $$D(\rho, \sigma) = tr(P(\rho-\sigma))$$. We wish to prove a result for a probabilistic mixture of states, $$\sum_i p_i\rho_i$$ and $$\sum_i q_i\sigma_i$$.

There exists a projector $$P$$ such that

\begin{align} D\left(\sum_i p_i\rho, \sum_i q_i\sigma\right) &= \sum_i p_i tr(P\rho_i) - \sum_i q_i tr(P\sigma_i) \\ &=\sum_i p_i tr(P(\rho_i - \sigma_i)) + \sum_i (p_i - q_i) tr(P\sigma_i) \\ &\leq \sum_i p_i D(\rho_i,\sigma_i) + D(p_i,q_i) \end{align}

Here, $$D(p_i,q_i)$$ is the classical probability distance given by $$D(p_i,q_i) = \frac{1}{2}\sum_i |p_i - q_i|$$ for any pair of probability vectors.

I don't understand the very last step, particularly the second term and how $$tr(P\sigma_i)$$ disappeared. Why is $$D(p_i,q_i) = \frac{1}{2}\sum_i |p_i - q_i| \geq \sum_i (p_i - q_i) tr(P\sigma_i)$$?

What you are missing is that $$D(X,Y)=\max_P \mathrm{tr}|P(X-Y)|\ ,$$ where the maximum is over all projectors $$P$$.
Then, the inequality follows since you are replacing one maximization -- the whole expression is maximized over one P -- by many optimizations over independent $$P$$'s, which thus can take more optimal values (together with the triangle inequality): $$\max_P \mathrm{tr}|P(\sum X_i)| \le \sum \max_P \mathrm{tr}|PX_i|\ ,$$ with $$X_i$$ all the terms being summed in the second line.
In addition, it is used that $$\mathrm{tr}[P\sigma_i]\le \mathrm{tr}[\sigma_i]=1$$.
Let me add that this variational characterization of the trace norm -- $$\|X\|_1=\mathrm{tr}|X| = \max_P \mathrm{tr}|PX|$$ over all projectors $$P$$ -- is extremely useful, so it is good to keep it in mind. It is a special case of the also very useful min-max-theorem.