Answer
a) Negative.
b) $ p=q B_0\left[\dfrac{L^2+d^2}{2d}\right] $
Work Step by Step
a) According to the right-hand rule and since the particle moves up, the particle is negatively charged.
Therefore, $q$ is negative.
b)
We know that the momentum of a particle is given by
$$p=mv\tag 1$$
We also know that the particle was moving horizontally, then entered a magnetic field that exerts a force on it.
We also know that the magnetic field forces the particle to move on a circular path (for a small distance which is an arc) which means that the net force exerted on the electron is given by
$$\sum F=F_B=\dfrac{mv^2}{r}\tag 2$$
whereas $r$ is the radius of curvature, as you see in the figure below.
We know that the magnetic force is given by
$$F_B=qvB_0$$
Plugging into (2);
$$ qvB_0=\dfrac{mv^2}{r} $$
$$ q B_0=\dfrac{\overbrace{mv}^{p} }{r} $$
Hence,
$$p=q B_0r\tag 3$$
Now we need to find $r$ in terms of $d$ and $L$.
From the geometry of the figure below, and by applying Pythagorean therom
$$r^2=L^2+(r-d)^2=L^2+r^2-2rd+d^2$$
Thus,
$$0=L^2- 2rd+d^2$$
Solving for $r$;
$$r=\dfrac{L^2+d^2}{2d}$$
Plugging into (3);
$$\boxed{p=q B_0\left[\dfrac{L^2+d^2}{2d}\right]}$$