ABCDEF is regular hexagon  M Is The mid-point of $\overrightarrow{\mathrm{DE}}$.N is the mid-point of $\overrightarrow{\mathrm{AM}}$, P is the mid-point of $\overrightarrow{\mathrm{BC}}$.By resolving the vector $\overrightarrow{\mathrm{NP}}$ into components with respect to the vectors $\overrightarrow{\mathrm{AB}}$ and $\overrightarrow{\mathrm{AF}}$ is 

Let $\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{b}}$.  Then $\overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}$ also $\overrightarrow{\mathrm{AB}}=2\overrightarrow{\mathrm{BC}}=2\overrightarrow{\mathrm{b}}$in  $\mathrm{\Delta }$le ACD We have $\overrightarrow{\mathrm{AC}}+\overrightarrow{\mathrm{CD}}=\overrightarrow{\mathrm{AD}}$
$\begin{array}{l}\Rightarrow \overrightarrow{\mathrm{CD}}=\overrightarrow{\mathrm{AD}}-\overrightarrow{\mathrm{DC}}\\ \Rightarrow \overrightarrow{\mathrm{CD}}=2\overrightarrow{\mathrm{b}}-(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})\\ \Rightarrow \overrightarrow{\mathrm{CD}}=\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}\end{array}$
In  $\mathrm{\Delta }$le BCD we have
$\begin{array}{l}\overrightarrow{\mathrm{BD}}=\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CD}}=\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}\\ \overrightarrow{\mathrm{BD}}=2\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}\\ \therefore \overrightarrow{\mathrm{AE}}=2\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}\end{array}$
Now M is the mid-point of DE
$\begin{array}{l}\therefore \overrightarrow{\mathrm{AM}}=\frac{1}{2}(\overrightarrow{\mathrm{AE}}+\overrightarrow{\mathrm{AD}})\\ =\frac{1}{2}(2\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}+2\overrightarrow{\mathrm{b}})\\ =2\overrightarrow{\mathrm{b}}-\frac{1}{2}\overrightarrow{\mathrm{a}}\end{array}$
N is the mid point of AM
$\begin{array}{l}\therefore \overrightarrow{\mathrm{AN}}=\frac{1}{2}\overrightarrow{\mathrm{AM}}\\ =\frac{1}{2}\left(2\overrightarrow{\mathrm{b}}-\frac{\overrightarrow{\mathrm{a}}}{2}\right)\\ =\overrightarrow{\mathrm{b}}-\frac{1}{4}\overrightarrow{\mathrm{a}}\end{array}$
Now,
$\begin{array}{l}\overrightarrow{\mathrm{AN}}+\overrightarrow{\mathrm{NP}}=\overrightarrow{\mathrm{AP}}\\ \Rightarrow \overrightarrow{\mathrm{NP}}=\overrightarrow{\mathrm{AP}}-\overrightarrow{\mathrm{AN}}\\ \Rightarrow \overrightarrow{\mathrm{NP}}=\frac{1}{2}(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{AC}})-\overrightarrow{\mathrm{AN}}\end{array}$
($\because$p is the mid –point of BC)
$\begin{array}{l}\Rightarrow \overrightarrow{\mathrm{NP}}=\frac{1}{2}(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})-(\overrightarrow{\mathrm{b}}-\frac{1}{4}\overrightarrow{\mathrm{a})}\\ \Rightarrow \overrightarrow{\mathrm{NP}}=\frac{5}{4}\overrightarrow{\mathrm{a}}-\frac{1}{2}\overrightarrow{\mathrm{b}}\end{array}$
We have 
$\begin{array}{l}\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{a}}\text{ and }\overrightarrow{\mathrm{AF}}=\overrightarrow{\mathrm{CD}}=\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}\\ \therefore \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{AF}}+\overrightarrow{\mathrm{a}}\\ \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{AF}}+\overrightarrow{\mathrm{AB}}\end{array}$
Substituting  these in 1 we get
$\begin{array}{l}\overrightarrow{\mathrm{NP}}=\frac{5}{4}\overrightarrow{\mathrm{AB}}-\frac{1}{2}(\overrightarrow{\mathrm{AF}}+\overrightarrow{\mathrm{AB}})\\ \Rightarrow \overrightarrow{\mathrm{NP}}=\frac{3}{4}\overrightarrow{\mathrm{AB}}-\frac{1}{2}\overrightarrow{\mathrm{AF}}\end{array}$