The mean of 1, 3, 4, 5, 7, 4 is m. The numbers 3, 2, 2, 4, 3, 3,

$1,3,4,5,7,4$

Mean $=\frac{1+3+4+5+7+4}{6}$

$=\frac{24}{6}$

$=4$

$M=4$

Consider the numbers 3, 2, 2, 4, 3, 3, p.

Mean $=\frac{3+2+3+4+3+3+p}{7}$

$\Rightarrow 7 \times(4-1)=17+p$

$\Rightarrow 21=17+p$

$\Rightarrow p=4$

Arranging the numbers 3, 2, 2, 4, 3, 3, 4 in ascending order, we have

2, 2, 3, 3, 3, 4, 4

Median $=\left(\frac{n+1}{2}\right)^{\text {th }}$ term

$q=\left(\frac{7+1}{2}\right)^{\text {th }}$ term

$=\left(\frac{8}{2}\right)^{\text {th }}$ term

$=4^{\text {th }}$ term

$\therefore q=3$

So,

$p+q=4+3$

$=7$

Hence, the correct option is (d).