Let $M$ be the midpoint of $BC.$ Let $Q$ be the intersection of $CP$ with $AM.$
Note that $\angle ABC = \frac12(180^\circ - 100^\circ) = 40^\circ$
and therefore
$$ \angle PBA = \angle ABC - \angle PBC = 40^\circ - 35^\circ = 5^\circ.$$
Since $Q$ is on the perpendicular bisector of $BC,$ triangle $\triangle BQC$
is isoceles with $BQ = CQ.$
Moreover,
\begin{align}
\angle QBC &= \angle QCB = 30^\circ,\\
\angle PBQ &= \angle PBC - \angle QBC = 35^\circ - 30^\circ = 5^\circ,\\
\angle PQB &= \angle QCB + \angle QBC = 60^\circ,\\
\angle PQA &= \angle MQC = 90^\circ - \angle QCB = 60^\circ.\\
\end{align}
In summary, $\angle PBA = \angle PBQ$ and $\angle PQA = \angle PQB.$
That is, the rays $BP$ and $QP$ are the angle bisectors of angles
$\angle ABQ$ and $\angle AQB$ of triangle $\triangle ABQ.$
The angle bisectors of a triangle are concurrent,
hence $AP$ is an angle bisector of $\angle BAQ.$
But $\angle BAQ = 50^\circ,$
so $\angle BAP = \frac12\angle BAQ = 25^\circ.$
To simplify derivation introduce the notation:
$$
BC=a,\quad AC=b,\quad AB=c,\\
\measuredangle{CAB}=\alpha,\quad
\measuredangle{ABC}=\beta,\quad
\measuredangle{BCA}=\gamma;\\
\measuredangle{ABP}=\alpha_B,\quad
\measuredangle{ACP}=\alpha_C,\\
\measuredangle{BAP}=\beta_A,\quad
\measuredangle{BCP}=\beta_C,\\\
\measuredangle{CAP}=\gamma_A,\quad
\measuredangle{CBP}=\gamma_B.
$$
Then the initial inequality can be written as:
$$\small
\frac{a(\sin\alpha_B+\sin\alpha_C)
+b(\sin\beta_A+\sin\beta_C)+c(\sin\gamma_A+\sin\gamma_B)}{a+b+c}\le1.\tag{*}
$$
which can be proved as follows:
$$\begin{align}\small
&\scriptsize\frac{a(\sin\alpha_B+\sin\alpha_C)
+b(\sin\beta_A+\sin\beta_C)+c(\sin\gamma_A+\sin\gamma_B)}{a+b+c}\\
&\scriptsize=\frac{\sin\alpha(\sin\alpha_B+\sin\alpha_C)
+\sin\beta(\sin\beta_A+\sin\beta_C)+\sin\gamma(\sin\gamma_A+\sin\gamma_B)}
{\sin\alpha+\sin\beta+\sin\gamma}\tag1\\
&\scriptsize=\frac{(\sin\beta\sin\beta_A+\sin\gamma\sin\gamma_A)
+(\sin\alpha\sin\alpha_B+\sin\gamma\sin\gamma_B)
+(\sin\alpha\sin\alpha_C+\sin\beta\sin\beta_C)}
{\sin\alpha+\sin\beta+\sin\gamma}\tag2\\
&\scriptsize\le\frac{\sin\alpha+\sin\beta+\sin\gamma}{\sin\alpha+\sin\beta+\sin\gamma}=1\tag3.
\end{align}
$$
Explanation:
(1)
Substitution:
$$a=2R\sin\alpha, \quad b=2R\sin\beta, \quad c=2R\sin\gamma.$$
(2)
Rearrangement.
(3) Transformation of sine products to cosine differences and then from cosine sums to cosine products:
$$\begin{align}
&\scriptsize\sin\beta\sin\beta_A+\sin\gamma\sin\gamma_A\\
&\scriptsize=\frac{\cos(\beta-\beta_A)-\cos(\beta+\beta_A)}2
+\frac{\cos(\gamma-\gamma_A)-\cos(\gamma+\gamma_A)}2\\
&\scriptsize=\frac{\cos(\beta-\beta_A)+\cos(\gamma-\gamma_A)}2
-\frac{\cos(\beta+\beta_A)+\cos(\gamma+\gamma_A)}2\\
&\scriptsize=
\underbrace{\cos\frac{\beta+\gamma-\beta_A-\gamma_A}2}_{\ge0}
\underbrace{\cos\frac{\beta-\gamma-\beta_A+\gamma_A}2}_{\le1}
-
\underbrace{\cos\frac{\beta+\gamma+\beta_A+\gamma_A}2}_{=0}\cos\frac{\beta-\gamma+\beta_A-\gamma_A}2\tag4\\
&\scriptsize\le \cos\frac{\beta+\gamma-\alpha}2=\cos\left(\frac\pi2-\alpha\right)=\sin\alpha,
\end{align}
$$
where we used $\beta_A+\gamma_A=\alpha$.
There may arise an interesting question when does the inequality $(\text{*})$ degenerate to equality?
It is easily seen from $(4)$ that this happens if and only if the following equations hold:
$$
\begin {cases}
\beta-\beta_A=\gamma-\gamma_A\\
\gamma-\gamma_B=\alpha-\alpha_B\\
\alpha-\alpha_C=\beta-\beta_C
\end {cases},
$$
which together with
$$
\begin {cases}
\beta_A+\gamma_A=\alpha\\
\gamma_B+\alpha_B=\beta\\
\alpha_C+\beta_C=\gamma
\end {cases}
$$
implies:
$$
\begin {cases}
\beta_C=\gamma_B=\frac\pi2-\alpha\\
\gamma_A=\alpha_C=\frac\pi2-\beta\\
\alpha_B=\beta_A=\frac\pi2-\gamma
\end {cases}.
$$
Thus the equality holds if and only if the triangle $ABC$ is acute and $P$ is its circumcenter.
Note:
With the convention that the angle measures at a triangle vertex are signed and the direction towards the adjacent side is positive, the inequality (*) becomes valid for the whole plane. Correspondingly it degenerates to equality at the circumcenter of the triangle, regardless of its shape.
Best Answer
From your work we obtain: $$\cos\left(\measuredangle BAP'-\measuredangle BAC+\measuredangle CAP\right)-\cos\left(\measuredangle BAP'+\measuredangle BAC-\measuredangle CAP\right)=$$ $$=\cos\left(\measuredangle BAC-\measuredangle BAP'-\measuredangle CAP\right)-\cos\left(\measuredangle BAC-\measuredangle BAP'+\measuredangle CAP\right)$$ or $$\cos\left(\measuredangle BAP'+\measuredangle BAC-\measuredangle CAP\right)=\cos\left(\measuredangle BAC-\measuredangle BAP'+\measuredangle CAP\right)$$ or $$\sin\measuredangle BAC\sin(\measuredangle CAP-\measuredangle BAP')=0$$ and we are done!