1. First GP

   Let its common ratio be $r$.

   Given:

   • first term $= a$
   • third term $= b$

   So, $ar^2=b$ $r^2=\frac{b}{a}$

   The $11^{\text{th}}$ term is $T_{11}=ar^{10}=a(r^2)^5=a\left(\frac{b}{a}\right)^5=\frac{b^5}{a^4}$

2. Second GP

   Let its common ratio be $s$.

   Given:

   • first term $= a$
   • fifth term $= b$

   So, $as^4=b$ $s^4=\frac{b}{a}$

   The $p^{\text{th}}$ term is $T_p=as^{p-1}$

   We are given that this equals the $11^{\text{th}}$ term of the first GP: $as^{p-1}=\frac{b^5}{a^4}$

3. Use $s^4=\dfrac{b}{a}$

   Since $\frac{b^5}{a^4}=a\left(\frac{b}{a}\right)^5=a(s^4)^5=as^{20}$

   Therefore, $as^{p-1}=as^{20}$

   As $a>0$, cancel $a$: $s^{p-1}=s^{20}$

   Since $a$ and $b$ are distinct positive real numbers, $\dfrac{b}{a}\neq 1$, hence $s\neq 1$. Therefore exponents must be equal: $p-1=20$ $p=21$

4. Check options

   • A: $20$ ✗
   • B: $24$ ✗
   • C: $21$ ✓
   • D: $25$ ✗

Hence, the correct answer is C.