A

Step 1: Analyzing the tangential acceleration
The speed of the particle is given by $v = 2t^2$. The tangential acceleration ($a_t$) is the time derivative of the speed:
$$a_t = \frac{dv}{dt} = \frac{d(2t^2)}{dt} = 4t.$$
The tangential acceleration increases with time, so it does not match (p) but matches with the increasing nature in (q). Hence, (a) does not pair with (p).

Step 2: Analyzing centripetal acceleration
The centripetal acceleration ($a_c$) is given by the formula:
$$a_c = \frac{v^2}{R} = \frac{(2t^2)^2}{R} = \frac{4t^4}{R}.$$
This acceleration increases with time due to the $t^4$ term, thus matching (b) with (q).

Step 3: Analyzing angular speed
The angular speed ($\omega$) can be calculated from the relationship:
$$\omega = \frac{v}{R} = \frac{2t^2}{R}.$$
As $t$ increases, angular speed also increases; thus, (c) does not match to (r) but to (q).

Step 4: Analyzing the angle between total and centripetal acceleration
The total acceleration is the vector sum of the tangential and centripetal accelerations. As the tangential component increases and centripetal component relies on $v^2$, the angle between these two accelerations changes and it depends on R hence matching (d) with (s).

Final matches:
(a) does not match any,
(b) matches with (q),
(c) does not match with any,
(d) matches with (s). Therefore, the correct pairs are:
(a) - (p), (b) - (q), (c) - (r), (d) - (s). Thus our answer is:

Therefore, the correct answer is: A.