When it comes to solving any collision problem, it is always important to apply conservation of linear momentum and conservation of angular momentum. This gives us the first two equations.
Conservation of linear momentum
For two rigid bodies 1 and 2, the conservation of linear momentum states the following:
$$m_1 \mathbf{u}_1+ m_2 \mathbf{u}_2 = m_1 \mathbf{v}_1 + m_2 \mathbf{v}_2 $$
$m_i$ is the mass of body $i$, $\mathbf{u}_i$ is the velocity vector of body $i$ just before the collision, and $\mathbf{v}_i$ just after the collision.
Let body 1 be the projectile, and body 2 be the disc. Since the disc is initially stationary, we can simplify this to:
$$m_1 \mathbf{u}_1 = m_1 \mathbf{v}_1 + m_2 \mathbf{v}_2$$
This is the first equation we’ll need to solve the collision.
Conservation of angular momentum
For angular momentum about an arbitrary point $P$ to be conserved, the following mess of an equation should be (briefly) observed:
$$\left(\mathbf{r}_1-\mathbf{r}_P\right)\times\left(m_1 \mathbf{u}_1\right)+I_1\omega_1\mathbf{k}+ \left(\mathbf{r}_2-\mathbf{r}_P\right)\times\left(m_2 \mathbf{u}_2\right)+ I_2\omega_2\mathbf{k} = \left(\mathbf{r}_1-\mathbf{r}_P\right)\times\left(m_1 \mathbf{v}_1\right)+I_1\Omega_1\mathbf{k}+ \left(\mathbf{r}_2-\mathbf{r}_P\right)\times\left(m_2 \mathbf{v}_2\right)+ I_2\Omega_2\mathbf{k} $$
$\mathbf{r}_i$ is the position vector of the centre of mass of body $i$ , $\mathbf{r}_P$ is the position vector of point $P$, $I_i$ is the moment of inertia about the axis of rotation for body $i$, $\omega_i\mathbf{k}$ is the angular velocity vector of body $i$ just before the collision, and $\Omega_i\mathbf{k}$ is the angular velocity vector of body $i$ just after the collision.
By setting $\mathbf{r}_P=\mathbf{r}_2$, noting that body 2 is initially stationary, and noting that body 1 is a particle and thus has no moment of inertia, the equation simplifies to:
$$\mathbf{r}\times\left(m_1 \mathbf{u}_1\right)= \mathbf{r}\times\left(m_1 \mathbf{v}_1\right)+ I_2\Omega_2\mathbf{k}$$
where $\mathbf{r}=\mathbf{r}_1-\mathbf{r}_2=-\sqrt{R^2-Y^2}\mathbf{i} + Y\mathbf{j}$.
Note that all the terms are parallel to the (out-of-plane) $z$-axis, i.e. parallel to $\mathbf{k}$. Therefore we can use the dot product to multiply all of the terms by $\mathbf{k}$ without loss of information from the equation:
$$m_1\left(\mathbf{r}\times\mathbf{u}_1\right)\cdot\mathbf{k}= m_1\left(\mathbf{r}\times\mathbf{v}_1\right)\cdot\mathbf{k} + I_2 \Omega_2$$
By noting that:
$$\left(\mathbf{a}\times\mathbf{b}\right)\cdot\mathbf{c}= \left(\mathbf{b}\times\mathbf{c}\right)\cdot\mathbf{a} = \left(\mathbf{c}\times\mathbf{a}\right)\cdot\mathbf{b} $$
we can rewrite the equation as:
$$m_1\mathbf{r^*}\cdot\mathbf{u}_1= m_1\mathbf{r^*}\cdot\mathbf{v}_1 + I_2 \Omega_2$$
where $\mathbf{r^*}=\mathbf{k}\times\mathbf{r}=-Y\mathbf{i}-\sqrt{R^2-Y^2}\mathbf{j}$
In this form, we now have the second of the equations we need to solve the collision problem.
The third and final equation
We need one more equation to solve this problem. This final equation will arise from the fact that the two bodies stick to one another.
((In your question, you specify that there is no energy loss as a result of the collision. However, it is not actually possible for the total kinetic energy of the system to be conserved if both bodies stick* after the collision. If the collision were perfectly elastic (no energy loss), both bodies must rebound from each other after colliding. For this reason, it is important that conservation of kinetic energy is not applied.
*If the bond that sticks one body to the other is not perfectly rigid (i.e. it is possible to pull apart the body apart as if there was a spring joining them), then it would be theoretically possible for mechanical energy to be conserved. However, this would result in complex oscillating behaviour between the two bodies, much like an undamped mass-spring system.))
If the two bodies are to stick post-collision, it important that velocities of the bodies at the point of contact are the same. This results in the following kinematic condition:
$$\mathbf{v}_1 = \mathbf{v}_2 + \Omega_2\mathbf{k}\times\mathbf{r}$$
This can be simplified to give the final equation:
$$\mathbf{v}_1 = \mathbf{v}_2 + \Omega_2\mathbf{r^*}$$
In summary
Three equations have been derived:
$$m_1 \mathbf{u}_1 = m_1 \mathbf{v}_1 + m_2 \mathbf{v}_2$$
$$m_1\mathbf{r^*}\cdot\mathbf{u}_1= m_1\mathbf{r^*}\cdot\mathbf{v}_1 + I_2 \Omega_2$$
$$\mathbf{v}_1 = \mathbf{v}_2 + \Omega_2\mathbf{r^*}$$
There are three unknowns:
$$\mathbf{v}_1 \quad \mathbf{v}_2 \quad \Omega_2$$
These are the velocity of the projectile, the velocity of the centre of the disc, and the angular speed of the disc, respectively, immediately after the collision.
By using vector algebra, it is possible to solve these equation for the three unknowns. Then it may be necessary to perform some substitutions, including:
$$m_1=M_P$$
$$m_2=M_D$$
$$I_2 = \frac{1}{2}M_D R^2$$
$$\mathbf{u}_1=V_P \mathbf{i}$$
