In one text, the derivation for perspective projection goes like this: 
If $x',y',z'$ represent any point along the projection line,and $x_{prp},y_{prp},z_{prp}$ are the projection reference point, then:
$$x'=x-(x-x_{prp})u$$
$$y'=y-(y-y_{prp})u$$
$$z'=z-(z-z_{prp})u$$
where $0\le u \le 1$
Now, finding the value of u from the third equation above and replacing in the first and second equations, we get at $z_vp$:
$$ x_p=x'= x.\frac{z_{vp}-z_{prp}}{z-z_{prp}}+x_{prp}.\frac{z-z_{vp}}{z-z_{prp}}$$ $$ y_p=y'= y.\frac{z_{vp}-z_{prp}}{z-z_{prp}}+y_{prp}.\frac{z-z_{vp}}{z-z_{prp}}$$
and there is no projection transformation matrix given.
In the second text, the derivation is using similar triangles:
comparing the similar triangles we get:
$$x_p=\frac{-x_c}{\frac{z_c}{d}}$$
$$y_p=\frac{y_c}{\frac{z_c}{d}}$$
and the projection transformation for homogeneous coordinates x,y,z is given by: $$ \begin{bmatrix} -1&&0&&0&&0\\ 0&&1&&0&&0\\ 0&&0&&1&&0\\ 0&&0&&\frac{1}{d}&&0 \end{bmatrix} $$
so, my questions are:
- How do I find the projection matrix of coordinates as found from derivation 1?
- Both are given in the perspective projection area of these different texts, are both transformations the same? If not, how are they different?
The two constructions are essentially the same. In the first, the reference point (more usually called the view point or camera center) is arbitrary and the image plane is $z=z_{\text{vp}}$, while in the second the camera is at the origin and the image plane is $z=d$. In solving for $u$ in the first construction you’re effectively using similar triangles, so if you place the reference point at the origin and set $d=z_{\text{vp}}-z_{\text{pvp}}$ you end up with the same equations... almost.
There’s an important difference between the two constructions: the second one also includes a reflection in the $y$-$z$ plane (the image $y$-axis). This is because we want a right-handed coordinate system for the image, but we’re “looking” at the image plane from the wrong side. Many authors place the image plane at $z=f$, $f\lt0$ instead (so that the camera is looking “down” at the scene) to avoid having to introduce this extraneous reflection.
Constructing a projection matrix from the first set of equations is fairly straightforward. The projected image of the point with homogeneous coordinates $(x,y,z,1)$ is $$\left({z_{\text{vp}}-z_{\text{prp}} \over z-z_{\text{prp}}}x+{z-z_{\text{vp}}\over z-z_{\text{prp}}}x_{\text{prp}},{z_{\text{vp}}-z_{\text{prp}} \over z-z_{\text{prp}}}y+{z-z_{\text{vp}}\over z-z_{\text{prp}}}y_{\text{prp}},z_{\text{vp}},1\right).$$ The projection matrix can’t depend on the point being projected, so just as is done to construct the matrix in the second derivation, multiply through by the denominator so that all of the coordinates appear in separate terms. Once you’ve done this, you can read off the projection matrix from the resulting coordinate vector: $$P = \begin{bmatrix} z_{\text{vp}}-z_{\text{prp}} & 0 & x_{\text{prp}} & -z_{\text{vp}}x_{\text{prp}} \\ 0 & z_{\text{vp}}-z_{\text{prp}} & y_{\text{prp}} & -z_{\text{vp}}y_{\text{prp}} \\ 0 & 0 & z_{\text{vp}} & -z_{\text{vp}}z_{\text{prp}} \\ 0&0&1&-z_{\text{pvp}}\end{bmatrix}.$$ You can obtain the same matrix by undoing the reflection in the second derivation (replace the $-1$ with a $1$) and concatenating it with two translations that correspond to translating the origin to the viewpoint and the translating back after projection. I’ll leave grinding through that computation to you.
Both this and the projection matrix in the second derivation produce a point in the three-dimensional scene space. To get a matrix that produces two-dimensional image coordinates instead, you would delete its third row.
We can also obtain $P$ directly by using a slightly different construction: the projected image of a point $\mathbf X$ is the intersection of the line through $\mathbf X$ and the camera center $\mathbf C$ with the image plane $\mathbf\pi$. This can be computed directly by using the Plücker matrix of the line: $$\mathbf X' = (\mathbf C\mathbf X^T-\mathbf X\mathbf C^T)\mathbf\pi.$$ Rearranging and factoring out $\mathbf X$, we get $$(\mathbf C\mathbf\pi^T-\mathbf C^T\mathbf\pi I_4)\mathbf X.$$ I’ll again leave it to you to verify that the parenthesized expression produces the matrix $P$ from above. This construction is completely general: it works for any image plane and viewpoint including points at infinity, which correspond to parallel projections (so-called affine cameras).