I would try LBP (or any other descriptor, based on what your images (and task, i.e. classes) actually are). To deal with different sizes, you can use Bag of Words encoding.
PCA does dimensional reduction by expressing $D$ dimensional vectors on an $M$ dimensional subspace, with $M<D.$ The vector itself can be written as a linear combination of $M$ eigenvectors, where the eigenvector is itself a unit vector that lives in the $D$ dimensional space.
Consider, for example, a two dimensional space which we reduce to one dimension using PCA. We find that the principal eigenvector is the unit vector that points equally in the positive $\hat{x}$ and $\hat{y}$ direction, i.e.
$$
\hat{v} = \frac{1}{\sqrt{2}} (\hat{x} + \hat{y}).
$$
In this case I'm using the hat ($\hat{x}$) symbol to indicate that it's a unit vector. You can think of this as a one-dimensional line going through a two-dimensional plane. In our reduced space, we can express any point $w$ in the two dimensional space as a one-dimensional (or scalar) value by projecting it onto the eigenvector, i.e. by calculating $w \cdot \hat{v}.$ So the point $(3,2)$ becomes $5/\sqrt{2},$ etc. But the eigenvector $\hat{v}$ is still expressed in the original two dimensions.
In general, we express a $D$ dimensional vector, $x,$ as a reduced $M$ dimensional vector $a$, where each component $a_i$ of $a$ is given by,
$$
a_i = \sum_j x_j V_{i j}
$$
where $V_{i j}$ is the $j$th component of the $i$'th eigenvector, and $i = 1, \dots, M$ and $j = 1, \dots, D.$ For that to work, the $i$th eigenvector must have $D$ components to take an inner product with $x$.
In your case, you can express a "reduced" vector of 200 components by taking the original image, a vector of 65025 components, and taking its inner product with each of the 200 images, each of which has 65025 components. Each inner product result is a component of your 200-dimensional vector. We expect each eigenvector to have the same number of dimensions as the original space. That is, we expect $M$ eigenvectors, each of which are $D$-dimensional.
Best Answer
Just a hint, after reading your comment. Each image (face) is represented as a stacked vector of length $N$. The different faces make up a dataset stored in a matrix $X$ of size $K\times N$. You might be confused about the fact that you use the PCA to obtain a set of eigenvectors (eigenfaces) $I = \{u_1, u_2, \ldots, u_D\}$ of the covariance matrix $X^TX$, where each $u_i \in \mathbb{R}^{N}$. You don't reduce the number of pixels used to represent a face, but rather you find a small number of eigenfaces that span a space which suitably represents your faces. The eigenfaces still live in the original space though (they have the same number of pixels as the original faces).
The idea is, that you use the obtained eigenfaces as a sort of archetypes that can be used to perform face detection.
Also, purely in terms of storage costs, imagine you have to keep an album of $K$ faces, each composed of $N$ pixels. Instead of keeping all the $K$ faces, you just keep $D$ eigenfaces, where $D \ll K$, together with the component scores and you can recreate any face (with a certain loss in precision).