Given,

$A\left(3-i\right)\Rightarrow A\equiv \left(3,-1\right)$

$B\left(2+i\right)\Rightarrow B\equiv \left(2,1\right)$

Now,

$\left|z-3+i\right|=\left|z-2-i\right|$

$\Rightarrow \left|z-\left(3-i\right)\right|=\left|z-\left(2+i\right)\right|\equiv \left\{\left|z-A\right|=\left|z-B\right|\right\}$

$\Rightarrow \left|z-\left(3-i\right)\right|=\left|z-\left(2+i\right)\right|\equiv \left\{\left|P-A\right|=\left|P-B\right|\right\}$

Therefore, we get $AP=BP$.

Point $P$ is moving along the line $CM$ such that $AP=BP$.

Hence, locus of the point $P$ is the perpendicular bisector of line $AB$.