1. Given the angles in the triangle $\triangle ABC$ with point $P$ inside it, we have:
$$\angle BAP = 10^\circ, \quad \angle ABP = 20^\circ, \quad \angle PCA = 30^\circ, \quad \angle PAC = 40^\circ$$

2. First, we find $\angle APB$:
$$\angle APB = 180^\circ - \angle BAP - \angle ABP = 180^\circ - 10^\circ - 20^\circ = 150^\circ$$

3. Next, we find $\angle BPC$:
$$\angle BPC = 180^\circ - \angle APB - \angle PCA = 180^\circ - 150^\circ - 30^\circ = 100^\circ$$

4. Using the Law of Sines in $\triangle ABP$:
$$\frac{AP}{\sin 20^\circ} = \frac{BP}{\sin 10^\circ} = \frac{c}{\sin 150^\circ}$$
Since $\sin 150^\circ = \sin 30^\circ = \frac{1}{2}$, we have:
$$\frac{c}{\sin 150^\circ} = 2c$$
Therefore:
$$AP = 2c \sin 20^\circ, \quad BP = 2c \sin 10^\circ$$

5. Using the Law of Sines in $\triangle APC$:
$$\frac{AP}{\sin 30^\circ} = \frac{CP}{\sin 40^\circ}$$
Since $\sin 30^\circ = \frac{1}{2}$, we have:
$$\frac{AP}{\frac{1}{2}} = 2AP = \frac{CP}{\sin 40^\circ}$$
Therefore:
$$CP = 2AP \sin 40^\circ = 4c \sin 20^\circ \sin 40^\circ$$
Using the double-angle identity $\sin 2\theta = 2 \sin \theta \cos \theta$, we get:
$$CP = c (2 \cos 20^\circ - 1)$$

6. Let $\angle PBC = x$. Using the Law of Sines in $\triangle BPC$:
$$\frac{CP}{\sin x} = \frac{BP}{\sin (80^\circ - x)}$$
Substituting the values:
$$\frac{c (2 \cos 20^\circ - 1)}{\sin x} = \frac{2c \sin 10^\circ}{\sin (80^\circ - x)}$$
Simplifying, we get:
$$\frac{2 \cos 20^\circ - 1}{\sin x} = \frac{2 \sin 10^\circ}{\sin (80^\circ - x)}$$

7. Using the identity $\sin (80^\circ - x) = \cos (10^\circ + x)$, we have:
$$\frac{2 \cos 20^\circ - 1}{\sin x} = \frac{2 \sin 10^\circ}{\cos (10^\circ + x)}$$

8. After simplification and factorization, we get:
$$2 \cos 20^\circ \cos (10^\circ + x) = 2 \sin 10^\circ \sin x + \cos (10^\circ + x) = \cos (x - 10^\circ)$$

9. Solving the equation $\cos (30^\circ + x) = 0$:
$$30^\circ + x = 90^\circ \implies x = 60^\circ$$

The final answer is $\boxed{60^\circ}$