Well I did it basically with PoP and then Stewart to see why $BP<BQ$.

Simple Problem -

Attachments:
Sharygin_Geometry_Olympiad P6.pdf (121kb)

I actually used power of point theorem and stewarts theorem to do this problem(Problem 6) Let, $\overline{AB}=c,\overline{BC}=a,\overline{CA}=b$ and $\text{Pow}_\omega(X)$ denote the power of $X$ with respect to a circle $\omega$ Claim: $\text{Pow}_{(ABC)}(P)=\text{Pow}_{(ABC)}(Q)$ Proof: Since, $\overline{AP}=s-a$, $\overline{CP}=s-c$, $\overline{AQ}=s-c$ and $\overline{CQ}=s-a$, we have \begin{align*} \overline{AP}\cdot \overline{CP}=\overline{AQ}\cdot \overline{CQ} \end{align*}Then by Power of Point Theorem, we can say that \begin{align*} \text{Pow}_{(ABC)}(P)=\text{Pow}_{(ABC)}(Q) \quad \blacksquare \end{align*}Therefore, \begin{align*} \overline{BP}\cdot \overline{PP'}=\overline{AP}\cdot \overline{CP}=\overline{AQ}\cdot \overline{CQ}=\overline{BQ}\cdot \overline{QQ'}\\ \end{align*}Which implies that, \begin{align*} \boxed{\overline{BP}\cdot \overline{PP'}=\overline{BQ}\cdot \overline{QQ'}} \end{align*}Claim: $\overline{BQ}>\overline{BP}$ Proof: By applying Stewart Theorem in $\Delta ABC$ we have \begin{align*} \overline{BP}^2=\frac{(s-a)a^2+(s-c)c^2}{b}-(s-a)(s-c)\\ \overline{BQ}^2=\frac{(s-a)c^2+(s-c)a^2}{b}-(s-a)(s-c)\\ \overline{BQ}^2-\overline{BP}^2=\frac{(a-c)^2(a+c)}{b}>0 \end{align*}So we have $\overline{BQ}>\overline{BP}$. $\blacksquare$ If $\overline{BQ}>\overline{BP}$, therefore from equation (1) we will have $\overline{PP'}>\overline{QQ'}$ $\blacksquare$

Well, to be honest, it should have been specified that $BA\neq BC$; maybe the "respectively" part suggests it, but it's I don't think it's clearly stated. Anyhow, having assumed that $BA\neq BC$, here is my solution, including similar ideas to the ones posted above: Let's denote $a=BC$, $b=CA$, $c=AB$ and $s=\frac{a+b+c}{2}$. Let $\Gamma$ be the circumcircle of $\triangle ABC$, $\omega$ be the incircle of $\triangle ABC$ and $\Omega$ be the excircle of $\triangle ABC$ opposite point $B$ (we know that $Q$ lies on $AC$, so the excircle in the problem statement is the excircle of $\triangle ABC$ opposite point $B$ as the other two touch line $AC$ and not the segment). Let $\omega$ touch $AB$ at point $C_{1}$ and $BC$ at point $A_{1}$ and let $\Omega$ touch the line $AB$ at point $C_{2}$ and line $BC$ at $A_{2}$. It's well-known that $AQ=CP=s-c$, so also $CQ=b-AQ=b-(s-c)=s-a=b-CP=AP$. Now, by power of a point for point $Q$ in $\Gamma$ we have that: $$AQ\cdot QC = BQ\cdot QQ'$$And by power of a point for $P$ in $\Gamma$ we have that: $$AP\cdot PC=BP\cdot PP'$$However, we showed that $AQ=CP$ and $CQ=AP$, so: $$BQ\cdot QQ'=AQ\cdot QC=AP\cdot PC=BP\cdot PP'$$We will prove that $BP<BQ$. We'll use Stewart's theorem for $\triangle ABC$ for $P\in AC$ and $Q\in AC$. From Stewart's theorem for $\triangle ABC$ and $P,Q\in BC$ we have that: \[BP^2=\frac{PC\cdot AB^2+AP\cdot BC^2}{AC}-AP\cdot PC=\frac{(s-c)\cdot c^2+(s-a)\cdot a^2}{b}-(s-a)(s-c)\]\[BQ^2=\frac{AQ\cdot BC^2+CQ\cdot AB^2}{AC}-AQ\cdot CQ=\frac{(s-c)\cdot a^2+(s-a)\cdot c^2}{b}-(s-c)(s-a)\]\[\Longrightarrow BQ^2-BP^2=\frac{(s-c)\cdot c^2+(s-a)\cdot a^2}{b}-\frac{(s-c)\cdot a^2+(s-a)\cdot c^2}{b}=\]\[=\frac{1}{b}\left(\left(\frac{a+b-c}{2}\right)c^2+\left(\frac{-a+b+c}{2}\right)a^2-\left(\frac{a+b-c}{2}\right)a^2-\left(\frac{-a+b+c}{2}\right)c^2\right)=\]\[=\frac{1}{b}(c^3-c^2a-ca^2+a^3)=\frac{(c-a)^2(c+a)}{b}>0\]We used the assumption that $c\neq a$ (otherwise $P\equiv Q$ because of symmetry, but we rejected this case) and also $b>0$ and $a+c>0$. $$\Longrightarrow BQ^2>BP^2\Longrightarrow BQ>BP\Longrightarrow PP'>QQ'$$We have that $BQ\cdot QQ'=BP\cdot PP'$, but we showed that $BQ>BP$, so $PP'>QQ'$ which is what was wanted in the problem statement.

Incenter lies on segment between $A$ and $A-\text{excenter},$ so $P$ lies on segment $HQ,$ where $AH$ is altitude. Thus $|AQ|>|AP|,$ and we are done by equality $$|AP|\cdot |PP'|=|BP|\cdot |CP|=|BQ|\cdot |CQ|=|AQ|\cdot |QQ'|.$$

Marinchoo wrote: Well, to be honest, it should have been specified that $BA\neq BC$; maybe the "respectively" part suggests it, but it's I don't think it's clearly stated. Anyhow, having assumed that $BA\neq BC$, here is my solution, including similar ideas to the ones posted above: I was also thinking of the same... Then I mailed them regarding this... They also said that, in this prob $BA\neq BC$..

Notice that by PoP this amounts to proving that $BP < CQ$ which by pythagoras amounts to proving that the $B$-Altitude,$P$ and $Q$ are in an order, which is trivial because $B$, the incenter and the $B$-excenter are in order and the former points are just the projection of the latter $\blacksquare$