If $$C$$ is the mid point of $$AB$$ and $$P$$ is any point outside $$AB$$, then

Take an arbitrary origin $$O$$ and denote the position vectors of the points by
$$\overrightarrow{OA}= \mathbf a,\; \overrightarrow{OB}= \mathbf b,\; \overrightarrow{OC}= \mathbf c,\; \overrightarrow{OP}= \mathbf p$$.

Because $$C$$ is the midpoint of $$AB$$, the midpoint formula gives
$$\mathbf c = \frac{\mathbf a + \mathbf b}{2}\; \; -(1)$$

Convert the required vectors to position-vector form:

$$\overrightarrow{PA}= \mathbf a-\mathbf p,\;\; \overrightarrow{PB}= \mathbf b-\mathbf p,\;\; \overrightarrow{PC}= \mathbf c-\mathbf p$$

Add $$\overrightarrow{PA}$$ and $$\overrightarrow{PB}$$:
$$\overrightarrow{PA}+\overrightarrow{PB}= (\mathbf a-\mathbf p)+(\mathbf b-\mathbf p)=\mathbf a+\mathbf b-2\mathbf p\; \; -(2)$$

Double $$\overrightarrow{PC}$$ and substitute $$\mathbf c$$ from (1):
$$2\overrightarrow{PC}=2(\mathbf c-\mathbf p)=2\mathbf c-2\mathbf p =2\left(\frac{\mathbf a+\mathbf b}{2}\right)-2\mathbf p =\mathbf a+\mathbf b-2\mathbf p\; \; -(3)$$

Expressions (2) and (3) are identical, so
$$\overrightarrow{PA}+\overrightarrow{PB}=2\overrightarrow{PC}$$

Hence Option A is correct, while Options B, C are not satisfied by the vectors.

Final answer: Option A which is: $$\overrightarrow{PA} + \overrightarrow{PB} = 2\overrightarrow{PC}$$