Step 1:
Let $z=x+iy$
$\large\frac{2z+1}{iz+1}=\frac{2(x+iy)+1}{i(x+iy)+1}$
$\qquad\quad=\large\frac{2x+1)+2iy}{ix-y+1}$
$\qquad\quad=\large\frac{2x+1)+2iy}{1-y)+ix}\times \big(\large\frac{(1-y)-ix}{(1-y)-ix}\big)$
$\qquad\quad=\large\frac{(2x+1)(1-y)+2xy+i(2y(1-y)-x(2x+1))}{(1-y)^2+x^2}$
Step 2:
$Im\bigg[\large\frac{2z+1}{iz+1}\bigg]$$=-2$
$\large\frac{2y(1-y)-x(2x+1)}{(1-y)^2+x^2}$$=-2$
$2y-2y^2-2x^2-x=-2[1+y^2-2y+x^2]$
$2y-2y^2-2x^2-x=-2-2y^2+4y-2x^2$
$x+2y=2$
The locus of $P$ is a straight line.