It's not necessarily an altitude!

We use barycentric coordinates.

As usual, we take $A=(1,0,0), B=(0,1,0)$ and $C=(0,0,1)$ .

The point $X=(0,\frac{1}{3}, \frac{2}{3})$ . So the equation of the line $AX$ is $z=2y$ . So point $P$ can be parametrised as

$P= (1-3a,a,2a)$ . Now the line $BP$ has equation $2ax = (1-3a)z$ , so it intersects the line $AC$ at the point

$Y = (1-3a:0:2a)= (\frac{1-3a}{1-a} , 0 , \frac{2a}{1-a})$ . Similarly, the point $Z = (\frac{1-3a}{1-2a},\frac{a}{1-2a},0)$ .

Let the equation of the circle through $B,Z,Y$ and $C$ be

$-a^2yz-b^2zx-c^2xy + ux+vy+wz=0$ .

Placing the coordinates of $B=(0,1,0)$ , we get $v=0$ . Placing the coordinates of $C=(0,0,1)$ , we get $w=0$ . Now we place the coordinates of $Y$ into the equation. Because $y_Y = 0$ , so

$-b^2z_Yx_Y + ux_Y =0$ , and because $a \neq 1$ (Why?), $x_Y \neq 0$ . Thus

$u = b^2z_Y$ .

Similarly, by placing the coordinates of $Z$ into the equation, we get

$u = c^2y_Z$ . It follows that

$b^2\frac{2a}{1-a} = c^2\frac{a}{1-2a}$ . Since $\frac{c}{b} = \frac{4}{3}$ , we get

$\frac{18a}{1-a} = \frac{16a}{1-2a}$ , or $a = \frac{1}{10}$ (again, why is $a \neq 0$ ?)

Thus $\frac{AP}{PX} = \frac{x_A-x_P}{x_P-x_X} = \frac{3a}{1-3a} = \frac{\frac{3}{10}}{1-\frac{3}{10}} = \boxed{\frac{3}{7}}$ . Therefore $a=3, b=7,$ and $\boxed{a+b=10}$ .

For this solution, we will utilize Ceva's Theorem and van Aubel's Theorem .

Firstly, we can apply Ceva's Theorem to get $\dfrac{XB}{XC} \cdot \dfrac{YC}{YA} \cdot \dfrac{ZA}{ZB} = 1.$ We also know that $\dfrac{ZA}{YA} = \dfrac{AC}{AB},$ since $BZYC$ is cyclic. Combining these with the given statements, we have

$\begin{aligned} \dfrac{XB}{XC} \cdot \dfrac{YC}{YA} \cdot \dfrac{ZA}{ZB} &= 1 \\ 2 \cdot \dfrac{YC}{YA} \cdot \dfrac{ZA}{ZB} &= 1 \\ \dfrac{YC}{YA} \cdot \dfrac{ZA}{ZB} &= \dfrac{1}{2} \\ \dfrac{YC}{ZB} \cdot \dfrac{ZA}{YA} &= \dfrac{1}{2} \\ \dfrac{YC}{ZB} \cdot \dfrac{AC}{AB} &= \dfrac{1}{2} \\ \dfrac{YC}{ZB} \cdot \dfrac{3}{4} &= \dfrac{1}{2} \\ \dfrac{YC}{ZB} &= \dfrac{2}{3}. \end{aligned}$

Now, we focus on $\triangle BZP$ and $\triangle CPY.$ Because $\angle BZP = \angle PYC$ and $\angle ZBP = \angle PCY$ due to the cyclic nature of $BZYC,$ we get $\triangle BZP \sim \triangle CPY.$ This means

$\dfrac{YC}{ZB} = \dfrac{CP}{BP} = \dfrac{PY}{PZ} = \dfrac{2}{3},$

which we had found previously.

By van Aubel's Theorem,

$\dfrac{CP}{PZ} = \dfrac{CX}{XB} + \dfrac{CY}{YA} = \dfrac{1}{2} + \dfrac{CY}{YA} \\ \dfrac{BP}{PY} = \dfrac{BX}{XC} + \dfrac{BZ}{ZA} = 2 + \dfrac{BZ}{ZA}. \\$

Dividing the top equation by the bottom yields

$\begin{aligned} \dfrac{\frac{1}{2} + \frac{CY}{YA}}{2 + \frac{BZ}{ZA}} &= \dfrac{CP}{PZ} \cdot \dfrac{PY}{BP} \\ &= \dfrac{CP}{BP} \cdot \dfrac{PY}{PZ} \\ &= \dfrac{2}{3} \cdot \dfrac{2}{3} \\ &= \dfrac{4}{9}. \end{aligned}$

Let $a = \dfrac{CY}{YA}, b = \dfrac{BZ}{ZA}.$ The relationship above simplifies into, $\dfrac{1}{2} + a = \dfrac{4}{9} (2 + b).$ However, by Ceva's, we also know that $\dfrac{YC}{YA} \cdot \dfrac{ZA}{ZB} = \dfrac{1}{2},$ so $a = \dfrac{1}{2}b.$ Solving this system of equations yields $(a, b) = \left ( \dfrac{7}{2}, 7 \right ).$

Finally, applying van Aubel's Theorem again, we have

$\begin{aligned} \dfrac{AP}{PX} &= \dfrac{AZ}{ZB} + \dfrac{AY}{YC} \\ &= \dfrac{1}{b} + \dfrac{1}{a} \\ &= \dfrac{1}{7} + \dfrac{2}{7} \\ &= \dfrac{3}{7}. \end{aligned}$

Since we never made any assumptions about $\triangle ABC$ other than what was stated in the problem, we see that $\dfrac{AP}{PX} = \dfrac{3}{7}$ is the only solution that achieves our desired results, so $p + q = \boxed{10}.$