If the measurement is an irreversible process then the probability of
the resultant state is towards the initial quantum state it was in,
that is, if the resultant state is $1$ with a probability of, say,
$63\%$, that means that is has a $63\%$ probability that the initial
state was $|\psi \rangle= |1\rangle$.
This is a fundamental misinterpretation. If the resultant state after measurement is $|1\rangle$ with $63\%$ probability, then it simply means that the initial quantum state was of the form $\sqrt{0.27}e^{i\theta}|0\rangle + \sqrt{0.63}e^{i\phi}|1\rangle$ where $\theta,\phi\in \Bbb R$ (of course, assuming we measured in the $\{|0\rangle, |1\rangle\}$ basis). This initial quantum state collapses to the state $|1\rangle$ with $63\%$ probability! Don't think of it as "if the resultant state is $|1\rangle$ with $63\%$ probability it means that the initial quantum state was $|1\rangle$ with $63\%$ probability." That's wrong. Qubits can totally exist in a linear superposition of $|0\rangle$ and $|1\rangle$ and do not necessarily need to be either one of them i.e. either $|0\rangle$ or $|1\rangle$. A superposition is a perfectly valid state of existence for a qubit! It also does not make sense to say: it's both $|0\rangle$ and $|1\rangle$ at the same time. Quantum mechanics is a mathematical model of reality and it does not need to match with your classical intuitions and need not be describable with the limited English vocabulary we apply for classical objects. We need to extend our vocabulary to describe qubits and that's where mathematics helps us. The word "superposition" is a mathematical term in this context.
However, after it has collapsed to a bit, then, there is no
probability associated with it, it certainly is $1$ for all future
measurements, it shouldn't be a probabilistic bit.
I believe the word in the term that's confusing you is "converts". I'd say don't overthink this. They simply mean that on measurement of a qubit in the state $a|0\rangle + b|1\rangle$ there $|a|^2$ probability of getting $|0\rangle$ and $|b|^2$ probability of getting $|1\rangle$. The term "probabilitic bit" here is simply an abstraction. They mean that upon measurement you can imagine your qubit to be a "bit" with a value of either 0 or 1. That is, it has the value 0 with probability $|a|^2$ and value 1 with probability $|b|^2$. But this is assuming they don't know the measurement result yet. They're simply mathematically modelling the measured qubit as a probabilitic bit without the knowledge of the result of the measurement. Of course, if you know the result of the measurement it's going to be either 0 or 1 with $100\%$ probability.
In the circuit below, what is the need of registering the state of the
qubit in a classical bit if it has collapsed irreversibly into that
same state? Does it only make sense to do so when it is the end of the
circuit?
This is already answered in Why do we need a Classical Register for carrying out Quantum Computations?
The results of the measurements are basically classical information i.e. 0 or 1 if you're measuring in the $\{|0\rangle, |1\rangle\}$ basis. Remember that in reality you're measuring things like electron spin. You need to store that classical information somewhere for future use, and that is what copying the results of the measurement to the classical bits achieves. And yes, in the gate model it's generally done after the measurements (which is usually the final step of a quantum algorithm).
