School-level vector question for finding length ratios in a triangle https://revisionmaths.com/sites/mathsrevision.net/files/imce/Questionpaper-Paper1H-November2018.pdf

$OAB$ is a triangle. $OPM$ and $APN$ are straight lines. $M$ is the midpoint of $AB$ and $N$ is on segment $OB$.
$$\overrightarrow{OA} = \mathbf{a}, \overrightarrow{OB} = \mathbf {b}, OP : PM = 3 : 2$$
Work out the ratio $ON:NB$

(I attach the link with the question as it is geometric, and I'm not allowed to post images yet.)
I've been trying to solve question 21 in this past paper for a while, but, to my surprise, I keep running into the same issue.
The initial stages are straightforward: find the lengths $AM$, $MB$, $AP$, $OP$ and $PM$. However, the last leap necessary to find the ratio $ON:NB$ evades me. I introduce a scalar $k$ such that $\overrightarrow{ON} = k\mathbf{b}$ and then try to solve it that way (as the mark scheme suggests). I've also tried introducing another scalar $q$ such that $\overrightarrow{AN} = q\overrightarrow{AP}$ looking for a pair of simultaneous equations to solve, but in the end the exact values I'm looking for always cancel out and I end up with $0 = 0$.
I must be missing something very obvious. How do you solve this question using the scalar?
P.S. – the mark scheme for this question offers no proper explanation of the solution beyond the obvious: https://revisionmaths.com/sites/mathsrevision.net/files/imce/Markscheme-Paper1H-November2018.pdf
 A: Michael showed you an alternative solution to the more geometric one (producing the parallelogram $OACB$ and similar triangles $ACP$, $NOP$) in the mark scheme.  But for the vector approach, we know
\begin{align*}
\overrightarrow{OM}&=\frac12(\mathbf{a}+\mathbf{b})\\
\overrightarrow{OP}&=\frac35\overrightarrow{OM}=\frac3{10}(\mathbf{a}+\mathbf{b})\\
\overrightarrow{AP}&=\frac3{10}\mathbf{b}-\frac7{10}\mathbf{a}\\
\overrightarrow{ON}&=k\,\mathbf{b}\\
\overrightarrow{AN}&=-\mathbf{a}+k\,\mathbf{b}
\end{align*}
But $\overrightarrow{AP}$ and $\overrightarrow{AN}$ are parallel, so the ratios of coefficient of $\mathbf{b}$ to coefficient of $\mathbf{a}$ must be the same:
$$
\frac{k}{-1}=\frac{3/10}{-7/10}
$$
so $k=3/7$ and $ON:NB=3:4$.
A: If you are solving using vectors,
Say, $\vec {ON} = k \textbf{b}$
$ \displaystyle  \vec {OP} = \frac{3}{5} \vec OM = \frac{3 (\textbf{a} + \textbf{b})}{10}$
$ \displaystyle \vec {AP} = \vec {OP} - \vec {OA} = \frac{- 7 \textbf{a} + 3 \textbf{b}}{10}$
$\vec {AN} = \vec {ON} - \vec {OA} =  - \textbf {a} + k \textbf{b}$
Now as $A, P, N$ are collinear, we must have
$\vec {AP} \times \vec {AN} = 0$
$(- 7 \textbf{a} + 3 \textbf{b}) \times ( - \textbf {a} + k \textbf {b}) = 0$
That leads to, $k = \cfrac{3}{7}$
So, $ON:NB = 3:4$
A: Let $K$ be a mid-point of $NB$.
Thus, $MK||AN$ and $ON:NK=OP:PM=3:2$, which says $ON:NB=3:4.$
