Expression of reflection isometry in the complex plane Using the fact that an anti-displacement in the plan has the form
$$f(z) = a \overline{z} + b$$ 
I have done some computation to find the reflection about the line passing through two points $P$ and $P'$. and I found that
$$R_{(PP')}(z) = \frac{P'-P}{\overline{P'-P}}(\overline{z} - \overline{P})+P $$
So ,I have two requests: The first is, I want to verify if this is truly the expression of the reflection of $z \in \mathbb{C}$  about the line passing through two points $P \in \mathbb{C}$ and $P' \in \mathbb{C}$.
And the second is, (if it is true)I want to find a reference where this formula can be found. 
 A: As a composition of primitive transformations
Let's build this up step by step.


*

*Move $P$ into the origin. So you want to reflect $z-P$ in the line spanned between the origin and $P'-P$.

*Normalize the direction “vector” of that line to $$\frac{P'-P}{\lvert P'-P\rvert}$$

*Rotate things so that that direction vector becomes the real axis. Since the number from step 2 has unit absolute value, this means either a division by that number, or a multiplication by its complex conjugate. So it maps $(z-P)$ to $$\frac{\bar P'-\bar P}{\lvert P'- P\rvert}(z-P)$$

*Now do the reflection, i.e. take the complex conjugate of everything. You end up at $$\frac{P'-P}{\lvert P'-P\rvert}(\bar z-\bar P)$$

*Undo the rotation from step 3, by multiplying with the vector from step 2. The squared absolute value can be used to cancel one factor. $$\left(\frac{P'-P}{\lvert P'-P\rvert}\right)^2(\bar z-\bar P)=
\frac{(P'-P)^2}{(P'-P)\overline{(P'-P)}}(\bar z-\bar P)=
\frac{\;P'-P\;}{\overline{P'-P}}(\bar z-\bar P)$$

*Undo the translation from step 1, by adding $P$, and you end up with your formula: $$\frac{\;P'-P\;}{\overline{P'-P}}(\bar z-\bar P)+P$$


So yes, your formula is correct. As for a reference, I don't know one, since I'd deduce this from a consideration like the one above.
As anti-möbius transformations
You could also approach this in another way. A reflection is a special case of Anit-Möbius transformation. As such it can be described by
$$z\mapsto \frac{a\bar z+b}{c\bar z+d}$$
for suitable $a,b,c,d\in\mathbb C$. This looks like four complex degrees of freedom, but since you can scale all your numbers by an arbitrary complex factor without changing the operation, you in fact only have three degrees of freedom. Which means that such an operation is uniquely defined by the images of three points. You can use three fixed points of your transformation for this, e.g. $P$, $P'$ and $(P+P')/2$.
A fixed point $z$ satisfies the equation
\begin{align*}
\frac{a\bar z+b}{c\bar z+d} &= z \\
a\bar z+b &= (c\bar z+d)z \\
a\bar z + b - c\lvert z\rvert^2 - dz &= 0
\end{align*}
(In my context, I'd use this as an operation on the Riemann sphere, operating on homogeneous coordinates, and take the point at infinity as the third fixed point. But that's just a matter of taste. With that setup, for $z=\infty\in\hat C$ the left hand side would be dominated by the term $c\lvert z\rvert^2$, from which you could directly conclude that $c=0$.)
Plugging in the three fixed points, you obtain a system of linear equations in $a,b,c,d$.
\begin{align*}
a\bar P + b - c\lvert P\rvert^2 - dP &= 0 \\
a\bar P' + b - c\lvert P'\rvert^2 - dP' &= 0 \\
2a\bar(P+P') + 4b - c\lvert P+P'\rvert^2 - 2d(P+P') &= 0
\end{align*}
As long as $P\neq P'$ it will have a one-dimensional space of solutions, from which you can pick any one you like. The one corresponding to your solution is likely the most readable one:
\begin{align*}
a &= (P'-P) &
b &= \overline{(P'-P)}P - (P'-P)\bar P = \bar P'P - P'\bar P \\
c &= 0 &
d &= \overline{(P'-P)}
\end{align*}
Note that this formulation looks a lot more symmetric than what you originally had, with $P$ and $P'$ playing almost the same role.
As anti-displacement
Of course, the anti-displacement you mentioned up front is just a special case of an Anti-Möbius transformation, where $c=0$. So large parts of this second section are not relevant to the question as posted, but might be useful to others. If you already know the form of the anti-displacement, then you can simply use two fixed points to compute $a$ and $b$ from that and be done.
I originally hadn't known the term anti-displacement, and can only now conclude that at least the formula you gave describes an Anti-Möbius transformation which fixes the point at infinity, i.e. a transformation which maps lines to lines, reverses angles and orientations but preserves the absolute value of angles. Or phrased differently, a combination of a reflection and a scaling, with arbitrary axis resp. center.
Doing the system of equations, you have
\begin{align*}
a\bar P + b &= P \\
a\bar P' + b &= P'
\end{align*}
or in matrix notation
$$
\begin{pmatrix} \bar P' & 1 \\ \bar P & 1 \end{pmatrix}\cdot
\begin{pmatrix}a\\b\end{pmatrix}=
\begin{pmatrix}P'\\P\end{pmatrix} \\
\begin{pmatrix}a\\b\end{pmatrix}=
\begin{pmatrix} \bar P' & 1 \\ \bar P & 1 \end{pmatrix}^{-1}\cdot
\begin{pmatrix}P'\\P\end{pmatrix}=
\frac1{\begin{vmatrix} \bar P' & 1 \\ \bar P & 1 \end{vmatrix}}
\begin{pmatrix} 1 & -1 \\ -\bar P & \bar P' \end{pmatrix}\cdot
\begin{pmatrix}P'\\P\end{pmatrix}=
\frac{1}{\;\overline{P'-P}\;}\begin{pmatrix}P'-P\\P\bar P'-P'\bar P\end{pmatrix}\\
a=\frac{P'-P}{\;\overline{P'-P}\;} \qquad
b=\frac{P\bar P'-P'\bar P}{\;\overline{P'-P}\;}
$$
Again this confirms your formula, although it takes some computation to see that these are indeed the same terms. The intermediate step from the previous section helps.
