How to get general projection of rect function?
Since your expression
$$\begin{align*}P(\theta,t) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \Pi(y)\Pi(x)\delta\left(x\cos\theta + y\sin\theta -t\right)dxdy\\
\end{align*}$$

uses a delta function with only 1 dimensional argument, in the 2 dimensional integration I assume you mean an infinite impulse line (an impulse "wall") and not a point impulse (an impulse "spike").

Let's just look at the general solution first, and then handle the degenrate case of $\theta = 0$.

I'll be using the following properties of the delta function:

$$\begin{align*}\delta(ax+b) &amp;= \dfrac{1}{|a|}\delta\left(x + \dfrac{b}{a}\right)\\
\\
\int_{-\infty}^{\infty} f(x)\delta(x-a)dx &amp;= f(a)\\
\end{align*}$$

So, I believe the solution is this:

$$\begin{align*}P(\theta,t) &amp;= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \Pi(y)\Pi(x)\delta\left(x\cos\theta + y\sin\theta -t\right)dxdy\\
\\
&amp;= \int_{-\infty}^{\infty} \Pi(y) \left[\int_{-\infty}^{\infty} \Pi(x)\dfrac{1}{|\cos\theta|}\delta\left(x - \left[-\dfrac{\sin\theta}{\cos\theta}y +\dfrac{t}{\cos\theta}\right]\right)dx\right]dy\\
\\
&amp;= \dfrac{1}{|\cos\theta|}\int_{-\infty}^{\infty} \Pi(y) \Pi\left(-\dfrac{\sin\theta}{\cos\theta}y +\dfrac{t}{\cos\theta}\right)dy\\
\\
&amp;\mbox{and if the rectangles have any overlap for the given $t$ and $\theta$}\\
\\
&amp;= \dfrac{1}{|\cos\theta|}\int_{\max\left(-\frac{1}{2},-\frac{1}{2}\frac{\cos\theta}{\sin\theta}+\frac{t}{\sin\theta}\right)}^{\min\left(\frac{1}{2},\frac{1}{2}\frac{\cos\theta}{\sin\theta}+\frac{t}{\sin\theta}\right)} dy\\
\\
&amp;= \dfrac{y}{|\cos\theta|}\biggr\rvert_{\max\left(-\frac{1}{2},-\frac{1}{2}\frac{\cos\theta}{\sin\theta}+\frac{t}{\sin\theta}\right)}^{\min\left(\frac{1}{2},\frac{1}{2}\frac{\cos\theta}{\sin\theta}+\frac{t}{\sin\theta}\right)}\\
\\
P(\theta,t) &amp;= \dfrac{\min\left(\frac{1}{2},\frac{1}{2}\frac{\cos\theta}{\sin\theta}+\frac{t}{\sin\theta}\right)-\max\left(-\frac{1}{2},-\frac{1}{2}\frac{\cos\theta}{\sin\theta}+\frac{t}{\sin\theta}\right)}{|\cos\theta|}
\end{align*}$$

So for the degenerate case of $\theta = 0$, the above expression runs into problems, but the integral for this special case is a bit easier:

$$\begin{align*}P(0,t) &amp;= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \Pi(y)\Pi(x)\delta\left(x\cos{0} + y\sin{0} -t\right)dxdy\\
\\
&amp;= \int_{-\infty}^{\infty} \Pi(y) \left[\int_{-\infty}^{\infty} \Pi(x)\delta\left(x-t\right)dx \right]dy\\
\\
&amp;= \Pi(t)\int_{-\infty}^{\infty} \Pi(y) dy\\
\\
&amp;= \Pi(t)
\end{align*}$$

The one dimensional Fourier Transform of $P(0,t)$ is a simple table look-up to find that it is a $\mathrm{sinc}()$ function.

The one dimensional Fourier Transform of the more general $P(\theta, t)$ is obviously more complicated.  There are 4 cases there.