The given equation for \(P\) is:
\[
P = P_0 \exp(-\alpha t^2)
\]
In this equation, \(P\) is a physical quantity, and \(P_0\) is the initial value of \(P\). The exponential function \(\exp(-\alpha t^2)\) is dimensionless, as the exponent of an exponential function must be dimensionless.
This means that the term inside the exponential must be dimensionless:
\[
-\alpha t^2
\]
Therefore, the dimensions of \(\alpha\) must cancel out the dimensions of \(t^2\), which means:
\[
[\alpha] = \frac{1}{[t^2]}
\]
So, \(\alpha\) has dimensions of \(t^{-2}\).
Thus, the correct answer is that \(\alpha\) has dimensions of \(t^{-2}\).