Solution

Let's define the $y$-axis perpendicular to bottom and the $x$-axis parallel to it. At $y$-axis there is no buoyancy force because between bottom and the cube there is no liquid.

The pressures at points $A$ and $C$ are

$$P_A = P + \rho_0 g(h+a\sin{\alpha})$$

$$P_C = P + \rho_0 gh$$

If both equations are rested,

$$P_A-P_C = \rho ga\sin{\alpha} \tag{1}$$

Note that difference of pressures between any point over side $AB$ and any point over $CD$ is expressed by $(1)$ (for any two points at same $y$ and at $a$ distance one from the other). If $(1)$ is multiplied by the side area $a^2$, buoyancy force can be found for $x$-axis

$$F_x = \rho g a^3 \sin{\alpha} \tag{2}$$

Applying Newton's Second Law for $x$-axis, also considering friction force

$$mg\sin{\alpha} = f_r + F_x$$

As $m = \rho a^3$ and according $(2)$

$$\boxed{f_r = a^3 g (\rho -\rho_0)\sin{\alpha}}$$

This force is "felt" by bottom in opposed direction of $x$-axis. Now, pressure force $F_p$ which acts in $y$-axis, generated by above water mass and the atmospher, must be found.

Atmospheric-pressure force is

$$F_0 = \frac{Pa^2}{\cos{\alpha}}$$

but the cube doesn't "feel" that force, else a component of it ($F_{0y}$).

$$F_{0y} = F_0\cos{\alpha} = Pa^2 \tag{3}$$

The pressure exerted by water mass is

$$F_w = m_w g \cos{\alpha} = \rho_0 V_w g \cos{\alpha} \tag{4}$$

This volume of water $V_w$ is

$$V_w = \frac{a^2h}{\cos{\alpha}} + \frac{a^3\tan{\alpha}}{2} \tag{5}$$

Putting $(5)$ into $(4)$

$$F_w = \rho_0 g a^3\left(\frac{h}{a}+\frac{\sin{\alpha}}{2}\right) \tag{6}$$

The pressure force over superior side of cube is

$$F_p = F_{0y} + F_w \tag{7}$$

Substituting $(3)$ and $(6)$ into $(7)$

$$F_p = Pa^2 + \rho_0 g a^3\left(\frac{h}{a}+\frac{\sin{\alpha}}{2}\right) \tag{8}$$

Applying Newton Second Law for $y$-axis

$$N = mg\cos{\alpha}+F_p = \rho a^3g\cos{\alpha}+F_p \tag{9}$$

and according to $(8)$

$$\boxed{N = a^3\rho_0 g \left(\frac{\rho}{\rho_0}\cos{\alpha}+\frac{1}{2}\sin{\alpha}+\frac{h}{a}\right)+Pa^2}$$

This normal force is directed in opposed direction of $y$-axis. Friction force $f_r$ is the parallel reaction force $F_{||}$ and normal force is the perpendicular one $F_{\perp}$.