1. Understand the geometry

   In parallelogram $ABCD$, $\overrightarrow{AB}=\vec q, \qquad \overrightarrow{AD}=\vec p.$

   The altitude from $B$ to the side $AD$ is the perpendicular from point $B$ to the line along $AD$.

   Since $AD$ has direction $\vec p$, the altitude vector from $B$ to $AD$ must:

   • start at $B$,
   • end on the line through $A$ in direction $\vec p$,
   • be perpendicular to $\vec p$.

2. Resolve $\vec q$ into components parallel and perpendicular to $\vec p$

   The projection of $\vec q$ on $\vec p$ is $\operatorname{proj}_{\vec p}(\vec q)=\frac{\vec p\cdot \vec q}{\vec p\cdot \vec p}\,\vec p.$

   Therefore, the component of $\vec q$ perpendicular to $\vec p$ is $\vec q-\frac{\vec p\cdot \vec q}{\vec p\cdot \vec p}\,\vec p.$

3. Direction of the altitude from $B$ to $AD$

   The vector found above is the perpendicular component of $\overrightarrow{AB}$, i.e. from the line $AD$ up to the point $B$.

   But the altitude is directed from $B$ to the side $AD$, so its direction is opposite.

   Hence, $\vec r=-\left(\vec q-\frac{\vec p\cdot \vec q}{\vec p\cdot \vec p}\,\vec p\right).$

   Simplifying, $\vec r=-\vec q+\frac{\vec p\cdot \vec q}{\vec p\cdot \vec p}\,\vec p.$

4. Match with the options

   This is exactly Option B: $\boxed{\vec r=-\vec q+\frac{\vec p\cdot \vec q}{\vec p\cdot \vec p}\,\vec p}.$

5. Check perpendicularity

   To verify: $\vec r\cdot \vec p=\left(-\vec q+\frac{\vec p\cdot \vec q}{\vec p\cdot \vec p}\vec p\right)\cdot \vec p$ $=-\vec q\cdot \vec p+\frac{\vec p\cdot \vec q}{\vec p\cdot \vec p}(\vec p\cdot \vec p)=-(\vec p\cdot \vec q)+ (\vec p\cdot \vec q)=0.$

   So $\vec r$ is indeed perpendicular to $AD$.

Therefore, the correct answer is B.