Edit: I had the rules wrong, below is a correct analysis with the correct rules.
To show how complicated this game is, let's try to solve a much simpler version. Suppose instead the deck has only 4 cards, namely, two kings and two queens, but all other rules are the same. Let's call the players Alice and Bob, with Alice going first.
If Alice is dealt either two queens or two kings, she immediately wins, so assume they are each dealt a queen and a king.
Alice has only two sensible options: she should play a single card, then either tell the truth (T) or lie (L) about its identity. Bob can then either doubt Alice's claim (D), play the card which Alice claimed to play (P), play the opposite card, or pass. If he passes, he immediately loses, and it turns out that if Bob can win dishonestly, then he can also win honestly, so we only consider the first two options.
TD: Bob is wrong, so Alice gets another turn, and wins.
LP: Alice just successfully lied. Bob plays a card, but then Alice plays her other card and wins.
TP: Alice plays one card, then Bob plays a card. Alice has no choice but to pass, as does Bob. Since Bob played last, he starts the next round and wins.
LD: Alice lied, but was called out, so she picks up her card, and Alice and Bob switch roles.
Let's say that the probability of Alice winning under optimal play is $p$. We can summarize the above information in a matrix, where each entry contains the probability of Alice winning in that situation:
$$
\begin{array}{c|c|c|}
&P&D\\\hline
T & 0&1\\\hline
L & 1 & 1-p\\\hline
\end{array}
$$
Notice the recursion above, because in the LD case, we have the same game with the roles reversed, so Alice now wins with probability $1-p$ instead of $p$.
The only solution is to use a mixed strategy. Suppose Alice tells the truth with probability $a$. Her best strategy is to choose $a$ so that Bob's payoff is equalized. If Bob chooses $P$, his expected payout is $1-a$, and if he chooses $D$, he gets $a\cdot 1+(1-a)(1-p)=1-p+pa$. Setting these equal, we get that
$$a = \tfrac{p}{1+p}$$
By symmetry, Bob will Believe with probability $ \frac{p}{1+p}$ as well.
Calculating Alice's probability of winning by summing over the four possibilities in the above matrix, then setting this equal to $p$, we get
$$
p = 2(\tfrac{p}{1+p})(\tfrac{1}{1+p})\cdot 1+(\tfrac{1}{1+p})^2(1-p)
$$
Amazingly, when you solve this, you get that $p$ is equal to the inverse of the Golden Ratio, i.e. $$p=\frac{\sqrt{5}-1}2\approx 61.8\%$$
is Alice's probability of winning under optimal play. To achieve this, she tells the truth about $\frac{.618}{1+.618}=38\%$ percent of the time. Bob should believe her with the same probability. Of course, this was assuming Alice was dealt a queen and a king. Overall, her probability of winning is $\frac13\cdot 1+\frac23\cdot 0.618=74.5\%$
We can perhaps guess, based on these results, that the game favors the first player, and the best strategy is to slightly lean towards deceit and doubt, but still be honest and trusting a nontrivial amount of the time.
However, you can see that the full 52 card game would be ridiculously complicated. This small game had itself as an option: imagine there were many games which mutually had each other as options. There would be a complicated web of nonlinear equations instead of a single nice one.
