Question:

1. Write the coordinates of the given points
\[ A(6,1),\; B(8,2),\; C(9,4),\; D(p,3). \]

2. Find the mid‑point of diagonal AC
\[ M_{AC}=\left(\frac{6+9}{2},\;\frac{1+4}{2}\right)=\left(\frac{15}{2},\;\frac{5}{2}\right)=(7.5,\;2.5). \]

3. Find the mid‑point of diagonal BD
\[ M_{BD}=\left(\frac{p+8}{2},\;\frac{3+2}{2}\right)=\left(\frac{p+8}{2},\;\frac{5}{2}\right) =\left(\frac{p+8}{2},\;2.5\right). \]

4. Equate the two mid‑points (since the diagonals bisect each other)
\[ \frac{p+8}{2}=7.5 \quad\text{and}\quad \frac{5}{2}=\frac{5}{2}. \]
The y‑coordinates are already equal; solve the x‑coordinate equation:
\[ \frac{p+8}{2}=7.5 \Rightarrow p+8=15 \Rightarrow p=7. \]

5. Conclusion
The value of \(p\) that makes \(ABCD\) a parallelogram is \(p=7\).