To check the principle of multiple proportions, a series of pure binary compounds (P$$_m$$Q$$_n$$) were analyzed and their composition is tabulated below. The correct option(s) is(are)

Compound	Weight % of P	Weight % of Q
1	50	50
2	44.4	55.6
3	40	60

First extract the data from the table given in the question.

Compound 1 → $$\%P = 50,\; \%Q = 50$$  ⇒ mass ratio $$\dfrac{P}{Q}= \dfrac{50}{50}=1$$
Compound 2 → $$\%P = 44.4,\; \%Q = 55.6$$  ⇒ mass ratio $$\dfrac{P}{Q}= \dfrac{44.4}{55.6}=0.8=\dfrac45$$
Compound 3 → $$\%P = 40,\; \%Q = 60$$  ⇒ mass ratio $$\dfrac{P}{Q}= \dfrac{40}{60}= \dfrac23$$

Case 1: Checking Option A - “If empirical formula of compound 3 is $$P_3Q_4$$, then empirical formula of compound 2 is $$P_3Q_5$$.”

For compound 3 assume $$P_3Q_4$$.
Weight ratio $$\dfrac{P}{Q}= \dfrac{3M_P}{4M_Q}= \dfrac23$$   $$\Rightarrow 9M_P = 8M_Q$$   $$\Rightarrow \dfrac{M_P}{M_Q}= \dfrac89$$ $$-(1)$$
With the same atomic masses, the ratio for $$P_3Q_5$$ (suggested for compound 2) would be
$$\dfrac{P}{Q}= \dfrac{3M_P}{5M_Q}= \dfrac{3\left(\tfrac89 M_Q\right)}{5M_Q}= \dfrac{24}{45}=0.533$$
The experimental ratio for compound 2 is $$0.8$$, not $$0.533$$. Hence Option A is WRONG.

Case 2: Checking Option B - “If empirical formula of compound 3 is $$P_3Q_2$$ and atomic weight of P is 20, atomic weight of Q is 45.”

Assume $$M_P = 20$$ and empirical formula $$P_3Q_2$$ for compound 3.
Mass of P in one empirical unit = $$3 \times 20 = 60$$
Let $$M_Q = x$$. Mass of Q in one empirical unit = $$2x$$.
Percentage of P: $$\dfrac{60}{60+2x}\times100 = 40\%$$ (from data of compound 3)
$$\dfrac{60}{60+2x}=0.40 \;\Rightarrow\; 60 = 0.40(60+2x)=24+0.8x$$
$$60-24 = 0.8x \;\Rightarrow\; 36 = 0.8x \;\Rightarrow\; x = 45$$
Thus $$M_Q = 45\;\text{u}$$. Option B is CORRECT.

Case 3: Checking Option C - “If empirical formula of compound 2 is $$PQ$$, then empirical formula of compound 1 is $$P_5Q_4$$.”

For compound 2 let empirical formula be $$PQ$$.
Weight ratio $$\dfrac{P}{Q}= \dfrac{M_P}{M_Q}=0.8=\dfrac45$$ ⇒ take $$M_P = 4k,\; M_Q = 5k$$ $$-(2)$$.
With these atomic masses, the ratio for $$P_5Q_4$$ (suggested for compound 1) becomes
$$\dfrac{P}{Q}= \dfrac{5M_P}{4M_Q}= \dfrac{5\,(4k)}{4\,(5k)}= \dfrac{20k}{20k}=1$$
This matches the experimental ratio of compound 1 (50 % : 50 % ⇒ 1 : 1). Hence Option C is CORRECT.

Case 4: Checking Option D - “If atomic weights of P and Q are 70 and 35 respectively, empirical formula of compound 1 is $$P_2Q$$.”

Given $$M_P = 70,\; M_Q = 35$$, compound 1 has equal masses of P and Q.
To get equal masses: number of moles $$n_P : n_Q = \dfrac{1}{70} : \dfrac{1}{35} = 1 : 2$$.
Thus the simplest integer formula is $$PQ_2$$, not $$P_2Q$$. Option D is WRONG.

Therefore the correct choices are:
Option B (atomic weight of Q is 45) and Option C (empirical formula $$P_5Q_4$$).

Final answer: Option B, Option C.