1. Let $A=3\sqrt{2} \quad \text{(point }(3\sqrt{2},0)\text{ on Argand plane)}$ and $B=p\sqrt{2}i \quad \text{(point }(0,p\sqrt{2})\text{)}.$

   Then the expression $|z-3\sqrt{2}|+|z-p\sqrt{2}i|$ represents the sum of distances of the point $z$ from fixed points $A$ and $B$.

2. For any point $z$ in the plane, the minimum value of $|z-A|+|z-B|$ is the distance between $A$ and $B$ itself, by triangle inequality. Hence $\min_z \left(|z-3\sqrt{2}|+|z-p\sqrt{2}i|\right)=|A-B|.$

3. Compute $|A-B|$: $A-B=3\sqrt{2}-p\sqrt{2}i.$ Therefore,

   =\sqrt{18+2p^2} =\sqrt{2(9+p^2)}.$$

4. Given that the minimum value is $5\sqrt{2}$, so $\sqrt{2(9+p^2)}=5\sqrt{2}.$

   Squaring both sides: $2(9+p^2)=50$ $9+p^2=25$ $p^2=16$ $p=\pm 4.$

5. From the options, the available value is $p=4.$

6. Checking options:

   • A: $3$ gives $p^2=9$, not valid
   • B: $\frac72$ gives $p^2=\frac{49}{4}$, not valid
   • C: $4$ gives $p^2=16$, valid
   • D: $\frac92$ gives $p^2=\frac{81}{4}$, not valid

Therefore the correct option is C.