1. **Problem statement:** Find the value of $k$ such that $\overrightarrow{AP} = \frac{3}{2} \overrightarrow{PB}$. 2. **Recall vector relation:** Since $P$ lies on line $AB$, we have $\overrightarrow{AP} = k \overrightarrow{AB}$. Also, $\overrightarrow{PB} = \overrightarrow{AB} - \overrightarrow{AP}$. 3. **Express $\overrightarrow{PB}$ in terms of $k$:** $$\overrightarrow{PB} = \overrightarrow{AB} - \overrightarrow{AP} = \overrightarrow{AB} - k \overrightarrow{AB} = (1 - k) \overrightarrow{AB}.$$ 4. **Set up the equation:** $$k \overrightarrow{AB} = \frac{3}{2} (1 - k) \overrightarrow{AB}.$$ 5. **Since $\overrightarrow{AB} \neq \vec{0}$, divide both sides by $\overrightarrow{AB}$:** $$k = \frac{3}{2} (1 - k).$$ 6. **Solve for $k$: $$k = \frac{3}{2} - \frac{3}{2}k$$ $$k + \frac{3}{2}k = \frac{3}{2}$$ $$\frac{5}{2}k = \frac{3}{2}$$ $$k = \frac{3/2}{5/2} = \frac{3}{5} = 0.6.$$ **Final answer:** $$k = \frac{3}{5}.$$