The argument is sound, but could use a bit of refinement. Below is how I might present it here on Math.SE. In a more formal setting, I'd likely be more verbose and less prone to fold explanations into the equations themselves ... perhaps to the detriment of comprehensibility.
Proof of Lemma. Drop perpendiculars from $P$ to $B'$ and $C'$ on lines $AB$ and $AC$, respectively; from $B$ to $H$ on line $AC$; and from $P$ to $Y$ on line $BH$. Note that $\square PC'HY$ is a rectangle. Also,
$$\angle B'BP \underbrace{\quad\cong\quad}_{\text{isos. $\triangle ABC$}} \angle ACB \underbrace{\quad\cong\quad}_{YP\parallel AC} \angle YPB$$
so that right triangles $\triangle PB'B$ and $\triangle BYP$ are congruent by Hypotenuse-(Acute)Angle. Therefore, we can write
$$\begin{align}
d(P,AB) \;+\; d(P,AC) &\;=\; \underbrace{|PB'|} \;+\;\;\; |PC'| \\
&\;=\; \;\;\color{blue}{|BY|}\;\; +\; \underbrace{|PC'|} &(\text{$\triangle PB'B\cong\triangle BYP$}) \\
&\;=\; \;\;|BY|\;\;+\;\;\;\color{red}{|YH|} &(\text{$\square PC'HY$ is a rectangle}) \\[6pt]
&\;=\;\;\;|BH| \;=\;d(B,AC)
\end{align}$$
Done! $\square$
Proof of Theorem. Consider point $P$ within equilateral $\triangle ABC$. Let the line through $P$ parallel to line $BC$ meet $AB$ and $AC$ at $B'$ and $C'$, respectively. Since both $\triangle AB'C'$ and $\triangle CAB$ are isosceles, we can simplify the sum of the distances from $P$ to the sides of $\triangle ABC$ thusly:
$$\begin{align}
\phantom{=}\quad \underbrace{d(P,BC)}\;&+\;d(P,AC)\;+\;d(P,AB) \\
=\quad \color{blue}{d(B',BC)}\;&+\;\underbrace{d(P,AC)\;+\;d(P,AB)} &(\text{$P$ and $B'$ are equidistant from $BC$})\\
=\quad d(B',BC)\;&+\;\qquad \color{red}{d(B',AC)} &(\text{Lemma applied to $P$ in isos. $\triangle AB'C'$})\\[6pt]
=\quad\quad\quad\;\; d(A&,BC) &(\text{Lemma applied to $B'$ in isos. $\triangle CAB$})
\end{align}$$
Done! $\square$
The use of underbraces and color may be a bit excessive (the latter also being problematic for color-blind readers), but I think they help, especially in the second string of arguments with all those $d(X,YZ)$ terms that might otherwise become a visual jumble. Even without those enhancements, simply using "displayed" equations makes the changes from one expression to the next easier to scan than using their in-line counterparts.
$$\begin{align}
\phantom{=}\quad d(P,BC)\;&+\;d(P,AC)\;+\;d(P,AB) \\[2pt]
=\quad d(B',BC)\;&+\;d(P,AC)\;+\;d(P,AB)\\[2pt]
=\quad d(B',BC)\;&+\;\qquad d(B',AC) \\[2pt]
=\quad\quad\quad\;\; d(A&,BC)
\end{align}$$
I'll note that I also incorporated extra space around the "$+$"s and "$=$"s, and some vertical padding, to improve on this rather dense alternative:
$$\begin{align}
\phantom{=} d(P,BC)&+d(P,AC)+d(P,AB) \\
=d(B',BC)&+d(P,AC)+d(P,AB) \\
=d(B',BC)&+\qquad d(B',AC) \\
=\quad\quad\; d(A&,BC)
\end{align}$$
Best Answer
Your proof looks okay to me, provided the notion that halves of congruent angles are congruent is in play. (At the foundations level, it can be tricky to know what's allowed, since different authors can present results in different orders.)
Note that your comment to MathMan suggests a way forward that's "safer" in this regard: Once you have $\triangle ADC\cong\triangle A'D'C'$, you can say $\angle A\cong\angle A'$ so that $\triangle ABC\cong\triangle A'B'C'$ by SAS. No subtle angle-algebra required.