It is true that:

We begin by recalling that the order of a reaction is defined exclusively from the experimentally obtained rate law, while a reaction mechanism tells us the actual sequence of elementary steps by which reactants are converted into products. The two ideas are related, but they are not the same; a given overall order may correspond to either a single elementary step or to many elementary steps, depending upon the kinetic details.

For an elementary (single-step) reaction the rate law can be written directly from the molecularity. If the step is $$\text{A} + \text{B} \longrightarrow \text{products},$$ then, by the law of mass action, the rate is $$\text{rate} = k[\text{A}][\text{B}],$$ which is second order overall. However, if the experimentally determined rate law of the overall process turns out to be second order, this does not compel us to accept that the reaction is elementary; the same rate law can arise after combining several elementary steps and applying the steady-state or pre-equilibrium approximations. Thus:

$$\text{“second-order”} \;\centernot\;\Rightarrow\;\text{“single step”}.$$

Next, when we meet a zero-order reaction, the experimental rate law is

$$\text{rate} = k[\text{A}]^0 = k,$$

which is independent of the concentration of the reactant. A constant rate that does not slow down as the reactants are consumed seldom originates from just one bimolecular or unimolecular collision. Instead, zero-order kinetics most commonly appear (i) in surface-catalysed processes where all active sites are saturated, or (ii) in photochemical reactions where the rate is controlled by a constant photon flux. In such cases several elementary events are involved: adsorption, surface reaction, desorption, or excitation and relaxation steps. Because at least two distinct molecular events are necessary, a zero-order overall rate law is essentially a consequence of a multi-step mechanism.

For a first-order reaction the empirical rate law is $$\text{rate} = k[\text{A}].$$ Many gas-phase isomerisations and certain radioactive decays are indeed single-step processes and first order, but the same first-order law can just as well arise from consecutive reactions in which the first step is slow and the later steps are fast. Therefore “first order” does not guarantee “single step”.

Lastly, declaring that “a zero-order reaction is a single step reaction” contradicts the discussion above, because a single elementary step that involves one or more reactant molecules must have an order equal to its molecularity (never zero).

Let us check each option one by one:

Option A: “A second order reaction is always a multistep reaction.” We have seen that a single elementary bimolecular collision already gives a second-order law, so the word “always” makes the statement false.

Option B: “A zero order reaction is a multi-step reaction.” Because zero-order kinetics usually stem from surface saturation or similar situations that necessarily involve more than one elementary step, this statement is true.

Option C: “A first order reaction is always a single step reaction.” Counter-examples exist (e.g. slow first step followed by rapid ones), so the statement is false.

Option D: “A zero order reaction is a single step reaction.” This is the converse of option B and is false for the same reasons.

Only option B survives scrutiny.

Hence, the correct answer is Option 2.