What is the value of the PL segment if QL = 12? For reference:

My progress...
the idea would be to demonstrate that $triangle AQL \cong \triangle LQC$
$\triangle PAB \sim QCB (AA)\\
\measuredangle PBA = \measuredangle QCB=90-\theta$

Could I say that $\triangle ABC$ is isoscelels without showing that BL is perpendicular to AC?
with the answer , the drawing to scale would look like this...


This is an interesting question. Please note your mistake is in trying to prove that $\triangle ABC$ is isosceles. $PL = QL$ holds regardless of whether $\triangle ABC$ is isosceles or not.
All we are given is that $\angle BAP = \angle BCQ = \theta$, and $BP$ and $BQ$ are perpendicular drawn to $AP$ and $CQ$ respectively. $L$ is the midpoint of $AC$.
We draw perp $BH$ from $B$ to $AC$. $LH = x$, $AL = LC = d$ (say).

As $\triangle BAP \sim \triangle BCQ$, $ \frac{BP}{BQ} = \frac{AP}{CQ} = \frac{AB}{BC} = k \ $ (say)
As quadrilateral $AHBP$ and $CHBQ$ are cyclic, we can apply Ptolemy's theorem.
In $AHBP$, $ \ AH \cdot BP + BH \cdot AP = AB \cdot HP$
$(d + x) \cdot k \cdot BQ + BH \cdot k \cdot CQ = k \cdot BC \cdot HP$
$(d + x) \cdot BQ + BH \cdot CQ = BC \cdot HP \tag1$
In $CHBQ$, $ \ CH \cdot BQ + BH \cdot CQ = BC \cdot HQ$
$(d - x) \cdot BQ + BH \cdot CQ = BC \cdot HQ \tag2$
Subtracting $(2)$ from $(1)$,
$2x \cdot BQ = BC \cdot (HP - HQ)$
$2x \sin \theta = HP - HQ \tag3$
Now $\angle BHP = \angle BAP = \theta$, so $\angle AHP = 90^\circ - \theta$. Similarly, $\angle BHQ = \angle BCQ = \theta$ and $\angle AHQ = 90^\circ - \theta$.
By law of cosine in $\triangle PHL$,
$PL^2 = x^2 + HP^2 - 2 x \cdot HP \cos(90^\circ - \theta)$
$PL^2 = x^2 + HP^2 - HP (HP - HQ) = x^2 + HP \cdot HQ\ $ (using $3$)
By law of cosine in $\triangle QHL$,
$QL^2 = x^2 + HQ^2 - 2 x \cdot HQ \cos(90^\circ + \theta)$
$QL^2 = x^2 + HQ^2 + HQ (HP - HQ) = x^2 + HP \cdot HQ\ $
So, $PL = QL = 12$
