If $$C$$ and $$D$$ are two events such that $$C \subset D$$ and $$P(D) \neq 0$$, then the correct statement among the following is:

The definition of conditional probability is

$$P(C \mid D)=\frac{P(C\cap D)}{P(D)}, \quad P(D)\neq 0.$$

Because $$C\subset D,$$ we have $$C\cap D=C.$$ Therefore

$$P(C \mid D)=\frac{P(C)}{P(D)}.$$

Since probabilities never exceed $$1,$$ we have $$P(D)\le 1.$$ Furthermore, when $$C\subset D,$$ the probability of $$C$$ cannot exceed that of $$D,$$ so $$0\le P(C)\le P(D)\le 1.$$

Dividing the same positive number $$P(C)$$ by a quantity $$P(D)$$ that is at most $$1$$ (and at least $$P(C)$$) makes the fraction at least as large as $$P(C):$$

$$\frac{P(C)}{P(D)}\ge P(C).$$

Hence

$$P(C \mid D)\ge P(C).$$

Equality holds only in the two boundary cases: (i) $$P(C)=0$$ (then both sides are $$0$$), or (ii) $$P(D)=1$$ (conditioning on the whole sample space does not change probability).

Therefore the correct statement is:

Option A which is: $$P(C \mid D) \geq P(C).$$